Rd Sharma 2021 Solutions for Class 9 Maths Chapter 5 Factorization Of Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Factorization Of Algebraic Expressions are extremely popular among Class 9 students for Maths Factorization Of Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 5.14:

#### Question 1:

*Factorize each of the following expressions:*

* p ^{3} + *27

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 2:

*y*^{3} + 125

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 3:

1 − 27*a*^{3}

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 4:

8*x*^{3}*y*^{3} + 27*a*^{3}

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 5:

64*a*^{3} − *b*^{3}

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 6:

$\frac{{x}^{3}}{216}-8{y}^{3}$

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 7:

10*x*^{4}*y* − 10*xy*^{4}

#### Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 8:

54*x*^{6}*y* + 2*x*^{3}*y*^{4}

#### Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 9:

32*a*^{3} + 108*b*^{3}

#### Answer:

The given expression to be factorized is

Take common from the two terms,. Then we have

This can be written in the form

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 10:

(*a* − 2*b*)^{3} − 512*b*^{3}

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 11:

8*x*^{2}*y*^{3} − x^{5}

#### Answer:

The given expression to be factorized is

Take common. Then we have

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 12:

1029 − 3x^{3}

#### Answer:

The given expression to be factorized is

Take common 3. Then we have from the above expression,

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 13:

*x*^{3}*y*^{3}^{ }+ 1

#### Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 14:

*x*^{4}*y*^{4} − *xy*

#### Answer:

The given expression to be factorized is

Take common. Then we have from the above expression,

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 15:

*a*^{3} + *b*^{3} + *a* +* b*

#### Answer:

The given expression to be factorized is

This can be written as

=

Recall the formula for sum of two cubes

Using the above formula, we have

Take common. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 16:

*Simplify*:

(i) $\frac{173\times 173\times 173+127\times 127\times 127}{173\times 173-173\times 127+127\times 127}$

(ii) $\frac{155\times 155\times 155-55\times 55\times 55}{155\times 155+155\times 55+55\times 55}\phantom{\rule{0ex}{0ex}}$

(iii) $\frac{1.2\times 1.2\times 1.2-0.2\times 0.2\times 0.2}{1.2\times 1.2+1.2\times 0.2+0.2\times 0.2}$

#### Answer:

(i) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for sum of two cubes

Using the above formula, the expression becomes

Note that both and *b* are positive. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

(ii) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both, *b* is positive and unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

(iii) The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both, *b* is positive and unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

#### Page No 5.14:

#### Question 17:

(*a* + *b*)^{3} − 8(*a* − *b*)^{3}

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 18:

(*x* + 2)^{3} + (*x* − 2)^{3}

#### Answer:

The given expression to be factorized is

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 19:

*x*^{6} + *y*^{6}

#### Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 20:

a^{12}+ b^{12}

#### Answer:

The given expression to be factorized is

This can be written as

Recall the formula for difference of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 21:

*x*^{3} + 6*x*^{2} + 12*x* + 16

#### Answer:

The given expression to be factorized is

This can be written as

Take common *x*^{2} from first two terms, 2*x* from the next two terms andfrom the last two terms. Then we have,

Finally, take common. Then we get,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 22:

${a}^{3}-\frac{1}{{a}^{3}}-2a+\frac{2}{a}$

#### Answer:

The given expression to be factorized is

This can be written as

Recall the formula for sum of two cubes

Using the above formula and taking common from the last two terms, we get

Take common. Then we have,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 23:

*a*^{3} + 3*a*^{2}*b** *+ 3*ab*^{2} + *b*^{3} − 8

#### Answer:

The given expression to be factorized is

Recall the well known formula

The given expression can be written as

Recall the formula for difference of two cubes

Using the above formula and taking common –2 from the last two terms, we get

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.14:

#### Question 24:

8*a*^{3} −* **b*^{3} − 4*ax* + 2*bx*

#### Answer:

The given expression to be factorized is

The given expression can be written as

Recall the formula for difference of two cubes

Using the above formula and taking common from the last two terms, we get

Take common. Then we have,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.17:

#### Question 1:

*Factorize:*

64*a*^{3} + 125*b*^{3} + 240*a*^{2}*b* + 300*ab*^{2}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.17:

#### Question 2:

125*x*^{3} − 27*y*^{3} − 225*x*^{2}*y* + 135*xy*^{2}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.17:

#### Question 3:

$\frac{8}{27}{x}^{3}+1+\frac{4}{3}{x}^{2}+2x$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.17:

#### Question 4:

8*x*^{3} + 27*y*^{3} + 36*x*^{2}*y* + 54*xy*^{2}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.17:

#### Question 5:

*a*^{3} − 3*a*^{2}*b* + 3*ab*^{2} −* **b*^{3} + 8

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the third and fourth terms. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

This can be written in the following form

Recall the formula for the sum of two cubes

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

#### Page No 5.17:

#### Question 6:

*x*^{3} + 8*y*^{3} + 6*x*^{2}*y* + 12*xy*^{2}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.18:

#### Question 7:

8*x*^{3} + *y*^{3} + 12*x*^{2}*y* + 6*xy*^{2}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common 6*xy* from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.18:

#### Question 8:

8*a*^{3}^{ }+ 27*b*^{3} + 36*a*^{2}*b* + 54*ab*^{2}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the sum of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.18:

#### Question 9:

8*a*^{3} − 27*b*^{3} − 36*a*^{2}*b *+ 54*ab*^{2}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common – 18*ab* from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

#### Page No 5.18:

#### Question 10:

*x*^{3} − 12*x*(*x* − 4) − 64

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common – 12*x* from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.18:

#### Question 11:

*a*^{3}*x*^{3} − 3*a*^{2}b*x*^{2} + 3*ab*^{2}*x* − *b*^{3}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common from the last two terms,. Then we get

This can be written in the following form

Recall the formula for the cube of the difference of two numbers

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.23:

#### Question 1:

*Factorize each of the following expressions:*

*a*^{3} + 8*b*^{3} + 64*c*^{3} − 24*abc*

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.23:

#### Question 2:

*x*^{3}^{ }− 8*y*^{3} + 27*z*^{3} + 18*xyz*

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis.

#### Page No 5.23:

#### Question 3:

27*x*^{3} − *y*^{3} − *z*^{3} − 9*xyz*

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

#### Page No 5.23:

#### Question 4:

$\frac{1}{27}{x}^{3}-{y}^{3}+125{z}^{3}+5xyz$

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

#### Page No 5.24:

#### Question 5:

8*x*^{3} +27*y*^{3} − 216*z*^{3} + 108*xyz*

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

#### Page No 5.24:

#### Question 6:

125 + 8*x*^{3} − 27y^{3} + 90*xy** *

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

#### Page No 5.24:

#### Question 7:

8*x*^{3} − 125*y*^{3} + 180*xy* + 216

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is ofis .

#### Page No 5.24:

#### Question 8:

Multiply:

(i) *x*^{2} + *y*^{2} + *z*^{2} − *xy* + *xz* + *yz* by *x* + *y* − *z*

(ii) *x*^{2} + 4*y*^{2} + *z*^{3} + 2*xy* + *xz* − 2*yz* by *x* − 2*y* − *z*

(iii) x^{2}^{ }+ 4y^{2} + 2*xy* − 3*x* + 6*y* + 9 by* x* − 2*y* + 3

(iv) 9*x*^{2} + 25*y*^{2} + 15*xy* + 12*x* − 20y + 16 by 3*x* − 5*y** *+ 4

(v) *x*^{2}^{ }+ 4*y*^{2} + z^{2} + 2*xy* + *xz* – 2*yz *by (−*z* + * x *– 2*y*)

#### Answer:

(v) *x*^{2}^{ }+ 4*y*^{2} + z^{2} + 2*xy* + *xz* – 2*yz *by (−*z* + *x *– 2*y*)

$\left({x}^{2}+4{y}^{2}+{z}^{2}+2xy+xz-2yz\right)\left(-z+x-2y\right)\phantom{\rule{0ex}{0ex}}=\left(-z\right)\left({x}^{2}+4{y}^{2}+{z}^{2}+2xy+xz-2yz\right)+\left(x\right)\left({x}^{2}+4{y}^{2}+{z}^{2}+2xy+xz-2yz\right)+\left(-2y\right)\left({x}^{2}+4{y}^{2}+{z}^{2}+2xy+xz-2yz\right)\phantom{\rule{0ex}{0ex}}=-{x}^{2}z-4{y}^{2}z-{z}^{3}-2xyz-x{z}^{2}+2y{z}^{2}+{x}^{3}+4x{y}^{2}+x{z}^{2}+2{x}^{2}y+{x}^{2}z-2xyz-2{x}^{2}y-8{y}^{3}-2y{z}^{2}-4x{y}^{2}-2xyz+4{y}^{2}z\phantom{\rule{0ex}{0ex}}={x}^{3}-8{y}^{3}-{z}^{3}-{x}^{2}z+2{x}^{2}y+{x}^{2}z-2{x}^{2}y-4{y}^{2}z+4x{y}^{2}-4x{y}^{2}+4{y}^{2}z-x{z}^{2}+2y{z}^{2}+x{z}^{2}-2y{z}^{2}-2xyz-2xyz-2xyz\phantom{\rule{0ex}{0ex}}={x}^{3}-8{y}^{3}-{z}^{3}-6xyz$

Hence, the required value is ${x}^{3}-8{y}^{3}-{z}^{3}-6xyz.$

#### Page No 5.24:

#### Question 9:

(3*x* − 2*y*)^{3} + (2*y** *− 4*z*)^{3} + (4*z* − 3*x*)^{3}

#### Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

,

when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis.

#### Page No 5.24:

#### Question 10:

(2*x* − 3*y*)^{3} + (4*z* − 2*x*)^{3} + (3*y* − 4*z*)^{3}

#### Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis .

#### Page No 5.24:

#### Question 11:

${\left(\frac{x}{2}+y+\frac{z}{3}\right)}^{3}+{\left(\frac{x}{3}-\frac{2y}{3}+z\right)}^{3}+{\left(-\frac{5x}{6}-\frac{y}{3}-\frac{4z}{3}\right)}^{3}$

#### Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and.

Then we have

We cannot further factorize the expression.

So, the required factorization is of is

#### Page No 5.24:

#### Question 12:

(*a* − 3*b*)^{3} + (3*b* − *c*)^{3} + (*c* − *a*)^{3}

#### Answer:

The given expression to be factorized is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the given expression can be written as

Put, and. Then we have

We cannot further factorize the expression.

So, the required factorization is ofis.

#### Page No 5.24:

#### Question 13:

$2{\sqrt{2a}}^{3}+3{\sqrt{3b}}^{3}+{c}^{3}-3\sqrt{6}abc$

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

#### Page No 5.24:

#### Question 14:

$3\sqrt{3}{a}^{3}-{b}^{3}-5\sqrt{5}{c}^{3}-3\sqrt{15}abc$

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is.

#### Page No 5.24:

#### Question 15:

$2\sqrt{2}{a}^{3}+16\sqrt{2}{b}^{3}+{c}^{3}-12abc$

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

We cannot further factorize the expression.

So, the required factorization is of is .

#### Page No 5.24:

#### Question 16:

(*x* – 2*y*)^{3} + (2*y *– 3*z*)^{3} + (3*z *– *x*)^{3}

#### Answer:

$\mathrm{To}\mathrm{solve}:{\left(x-2y\right)}^{3}+{\left(2y-3z\right)}^{3}+{\left(3z-x\right)}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{Let}a=\left(x-2y\right),b=\left(2y-3z\right),c=\left(3z-x\right).\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}a+b+c=\left(x-2y\right)+\left(2y-3z\right)+\left(3z-x\right)\phantom{\rule{0ex}{0ex}}=x-2y+2y-3z+3z-x\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{As}\mathrm{we}\mathrm{know}\mathrm{that},\phantom{\rule{0ex}{0ex}}\mathrm{if}a+b+c=0,\mathrm{then}{a}^{3}+{b}^{3}+{c}^{3}=3abc\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},{a}^{3}+{b}^{3}+{c}^{3}=3abc\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x-2y\right)}^{3}+{\left(2y-3z\right)}^{3}+{\left(3z-x\right)}^{3}=3\left(x-2y\right)\left(2y-3z\right)\left(3z-x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},{\left(x-2y\right)}^{3}+{\left(2y-3z\right)}^{3}+{\left(3z-x\right)}^{3}=3\left(x-2y\right)\left(2y-3z\right)\left(3z-x\right)$

#### Page No 5.24:

#### Question 17:

Find the value of *x*^{3} + *y*^{3} − 12*xy* + 64, when *x* + *y* =−4

#### Answer:

The given expression is

It is given that

The given expression can be written in the form

Recall the formula

Using the above formula, we have

#### Page No 5.24:

#### Question 18:

If *a, b, c* are all non-zero and *a + b + c* = 0, prove that $\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}=3$.

#### Answer:

$\mathrm{Given}:a+b+c=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{As}\mathrm{we}\mathrm{know}\mathrm{that},\phantom{\rule{0ex}{0ex}}\mathrm{if}a+b+c=0,\mathrm{then}{a}^{3}+{b}^{3}+{c}^{3}=3abc\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},{a}^{3}+{b}^{3}+{c}^{3}=3abc\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{by}\left(abc\right)\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{3}+{b}^{3}+{c}^{3}}{abc}=\frac{3abc}{abc}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{3}}{abc}+\frac{{b}^{3}}{abc}+\frac{{c}^{3}}{abc}=3\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ac}+\frac{{c}^{2}}{ab}=3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ac}+\frac{{c}^{2}}{ab}=3.$

#### Page No 5.25:

#### Question 1:

The factors of *x*^{3} −*x*^{2}*y* − *xy*^{2}^{ }+ *y*^{3} are

(a)(*x* + *y*) (*x*^{2} − *xy* + *y*^{2})

(b) (*x* + *y*) (*x*^{2} + *xy* + *y*^{2})

(c) (*x* + *y*)^{2} (*x* − *y*)

(d) (x − y)^{2} (x + y)

#### Answer:

The given expression to be factorized is

Take common from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

So, the correct choice is (d).

#### Page No 5.25:

#### Question 2:

The factors of x^{3} − 1 + y^{3} + 3xy are

(a) (*x* − 1 + *y*) (*x*^{2} + 1 +* **y*^{2} + *x* + *y* − *xy*)

(b) (*x* + *y* + 1) (*x*^{2} + *y*^{2} + 1 −*xy* − *x* −* y*)

(c) (*x* − 1 + *y*) (*x*^{2} − 1 − *y*^{2} + *x* + *y* + *xy*)

(d) 3(*x* + *y* −1) (*x*^{2} + *y*^{2} − 1)

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

So, the correct choice is (a).

#### Page No 5.25:

#### Question 3:

The factors of 8a^{3} + b^{3} − 6ab + 1 are

(a) (2*a* + *b *− 1) (4*a*^{2} + *b*^{2} + 1 − 3*ab* − 2*a*)

(b) (2*a* − *b* + 1) (4*a*^{2} + *b*^{2} − 4*ab* + 1 − 2*a* + *b*)

(c) (2*a* + *b *+ 1) (4*a*^{2} + *b*^{2} + 1 −2*ab* − *b* − 2*a*)

(d) (2*a* − 1 + *b*) (4*a*^{2} + 1 − 4*a* − *b* − 2*ab*)

#### Answer:

The given expression to be factorized is

This can be written in the form

Recall the formula

Using the above formula, we have

So, the correct choice is (c).

#### Page No 5.25:

#### Question 4:

(x + y)^{3} − (x − y)^{3} can be factorized as

(a) 2*y* (3*x*^{2} + *y*^{2})

(b) 2*x* (3*x*^{2} + *y*^{2})

(c) 2*y* (3*y*^{2} + *x*^{2})

(d) 2*x* (*x*^{2}+ 3*y*^{2})

#### Answer:

The given expression to be factorized is

Recall the formula for difference of two cubes

Using the above formula, we have,

So, the correct choice is (a).

#### Page No 5.25:

#### Question 5:

The expression (*a* − *b*)^{3} + (*b *− *c*)^{3} + (*c* −*a*)^{3} can be factorized as

(a) (*a* − *b*) (*b *− *c*) (*c* −*a*)

(b) 3(*a* − *b*) (*b* − *c*) (*c* −*a*)

(c) −3(*a* − *b*) (*b* −*c*) (*c* − *a*)

(d) (*a* + *b* + *c*) (*a*^{2} + *b*^{2} + *c*^{2} − *ab* − *bc* − *ca*)

#### Answer:

The given expression is

Let, and. Then the given expression becomes

Note that:

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

So, the correct choice is (b).

#### Page No 5.26:

#### Question 6:

The value of $\frac{(2.3{)}^{3}-0.027}{(2.3{)}^{2}+0.69+0.09}$

(a) 2

(b) 3

(c) 2.327

(d) 2.273

#### Answer:

The given expression is

This can be written in the form

Assumeand. Then the given expression can be rewritten as

Recall the formula for difference of two cubes

Using the above formula, the expression becomes

Note that both *a* and *b* are positive, unequal. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

So, the correct choice is (a).

#### Page No 5.26:

#### Question 7:

The value of $\frac{(0.013{)}^{3}+(0.007{)}^{3}}{(0.013{)}^{2}-0.013\times 0.007+(0.007{)}^{2}}$ is

(a) 0.006

(b) 0.02

(c) 0.0091

(d) 0.00185

#### Answer:

The given expression is

Assumeand. Then the given expression can be rewritten as

Recall the formula for sum of two cubes

Using the above formula, the expression becomes

Note that both and *b* are positive. So, neithernor any factor of it can be zero.

Therefore we can cancel the termfrom both numerator and denominator. Then the expression becomes

So, the correct choice is (b).

#### Page No 5.26:

#### Question 8:

*Mark the correct alternative in each of the following*:

The factors of *a*^{2} − 1 − 2*x* − *x*^{2} are

(a) (a − x + 1) (a − x − 1)

(b) (a + x − 1) (a − x + 1)

(c) (a + x +1) (a − x + 1)

(d) none of these

#### Answer:

The given expression to be factorized is

Take commonfrom the last three terms and then we have

So, the correct choice is (c).

#### Page No 5.26:

#### Question 9:

The factors of *x*^{4} + *x*^{2} + 25 are

(a) (*x*^{2} + 3*x* + 5) (*x*^{2} − 3*x* + 5)

(b) (x^{2}^{ }+ 3x + 5) (x^{2} + 3x − 5)

(c) (*x*^{2} + *x *+5) (*x*^{2} − *x* + 5)

(d) none of these

#### Answer:

The given expression to be factorized is

This can be written in the form

So, the correct choice is (a).

#### Page No 5.26:

#### Question 10:

The factors of *x*^{2} + 4*y*^{2} + 4*y* − 4*xy* − 2*x* − 8 are

(a) (*x *− 2*y* −4) (*x* − 2*y* + 2)

(b) (*x* − *y* + 2) (*x* − 4*y* − 4)

(c) (*x* + 2*y* − 4) (*x* + 2*y* + 2)

(d) none of these

#### Answer:

The given expression to be factorized is

This can be arrange in the form

Let. Then the above expression becomes

Put.

So, the correct choice is (a).

#### Page No 5.26:

#### Question 11:

The factors of *x*^{3} − 7*x* + 6 are

(a) *x* (*x* − 6) (*x *− 1)

(b) (*x*^{2} − 6) (*x* − 1)

(c) *(x* + 1) (*x* + 2) (*x* + 3)

(d) (*x *− 1) (*x* + 3) *(x* − 2)

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms. Then we have

Finally, take commonfrom the above expression,

So, the correct choice is (d).

#### Page No 5.26:

#### Question 12:

The expression *x*^{4} + 4 can be factorized as

(a) (*x*^{2} + 2*x** *+ 2) (*x*^{2} − 2*x* + 2)

(b) (*x*^{2} + 2*x* + 2) (*x*^{2} + 2*x* − 2)

(c) (*x*^{2} − 2*x* − 2) (*x*^{2}^{ }− 2*x* + 2)

(d) (*x*^{2} + 2) (*x*^{2} − 2)

#### Answer:

The given expression to be factorized is

This can be written in the form

So, the correct choice is (a).

#### Page No 5.26:

#### Question 13:

If 3*x* = *a* + *b* + *c*, then the value of (*x* − *a*)^{3} + (*x* −*b*)^{3} + (*x* − *c*)^{3} − 3(*x* − *a*) (*x* − *b*) (*x* −*c*) is

(a) *a* + *b* + *c*

(b) (*a* − *b*) (*b* − *c*) (*c* − *a*)

(c) 0

(d) none of these

#### Answer:

The given expression is

Recall the formula

Using the above formula the given expression becomes

Given that

Therefore the value of the given expression is

So, the correct choice is (c).

#### Page No 5.26:

#### Question 14:

If (*x* + *y*)^{3} − (*x* − *y*)^{3} − 6*y*(*x*^{2} − *y*^{2}) = *ky*^{3}, then *k* =

(a) 1

(b) 2

(c) 4

(d) 8

#### Answer:

The given equation is

Recall the formula

Using the above formula, we have

, provided.

So, the correct choice is (d).

#### Page No 5.26:

#### Question 15:

If *x*^{3} − 3*x*^{2} + 3*x* − 7 = (*x* + 1) (a*x*^{2} + *bx* + *c*), then *a* + *b* + *c* =

(a) 4

(b) 12

(c) −10

(d) 3

#### Answer:

The given equation is

*x*^{3} − 3*x*^{2} + 3*x* − 7 = (*x* + 1) (a*x*^{2} + *bx* + *c*)

This can be written as

${x}^{3}-3{x}^{2}+3x-7=\left(x+1\right)\left(a{x}^{2}+bx+c\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{3}-3{x}^{2}+3x-7=a{x}^{3}+b{x}^{2}+cx+a{x}^{2}+bx+c\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{3}-3{x}^{2}+3x-7=a{x}^{3}+\left(a+b\right){x}^{2}+\left(b+c\right)x+c$

Comparing the coefficients on both sides of the equation.

We get,

c = -7 .......(4)

Putting the value of *a* from (1) in (2)

We get,

So the value of *a, b *and *c* is 1, – 4 and -7 respectively.

Therefore,

*a* + *b* + *c* =1 - 4 - 7 = -10

So, the correct choice is (c).

#### Page No 5.27:

#### Question 1:

The factorized form of the expression *y*^{2} + (*x* – 1)*y *– *x* is ____________.

#### Answer:

${y}^{2}+\left(x-1\right)y-x\phantom{\rule{0ex}{0ex}}={y}^{2}+xy-y-x\phantom{\rule{0ex}{0ex}}=y\left(y+x\right)-1\left(y+x\right)\phantom{\rule{0ex}{0ex}}=\left(y-1\right)\left(y+x\right)$

Hence, the factorized form of the expression *y*^{2} + (*x* – 1)*y *– *x* is __( y â€‹– 1)(y + x)__.

#### Page No 5.27:

#### Question 2:

The factorized form of *a*^{3} + (*b* – *a*)^{3} – *b*^{3} is ____________.

#### Answer:

${a}^{3}+{\left(b-a\right)}^{3}-{b}^{3}\phantom{\rule{0ex}{0ex}}={a}^{3}+\left({b}^{3}-{a}^{3}-3ba\left(b-a\right)\right)-{b}^{3}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}{\left(x-y\right)}^{3}={x}^{3}-{y}^{3}-3xy\left(x-y\right)\right)\phantom{\rule{0ex}{0ex}}={a}^{3}+{b}^{3}-{a}^{3}-3ba\left(b-a\right)-{b}^{3}\phantom{\rule{0ex}{0ex}}=-3ba\left(b-a\right)\phantom{\rule{0ex}{0ex}}=3ab\left(a-b\right)$

Hence, the factorized form of *a*^{3} + (*b* – *a*)^{3} – *b*^{3} is __3 ab(a – b)__.

#### Page No 5.27:

#### Question 3:

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\mathrm{and}abc=2,\mathrm{then}a{b}^{2}{c}^{2}+{a}^{2}b{c}^{2}+{a}^{2}{b}^{2}c=\_\_\_\_\_\_\_\_\_\_\_\_.$

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1...\left(1\right)\phantom{\rule{0ex}{0ex}}abc=2...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}a{b}^{2}{c}^{2}+{a}^{2}b{c}^{2}+{a}^{2}{b}^{2}c\phantom{\rule{0ex}{0ex}}={a}^{2}{b}^{2}{c}^{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\phantom{\rule{0ex}{0ex}}={\left(abc\right)}^{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\phantom{\rule{0ex}{0ex}}={\left(2\right)}^{2}\left(1\right)\left(\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right)\phantom{\rule{0ex}{0ex}}=4$

Hence, if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\mathrm{and}abc=2,\mathrm{then}a{b}^{2}{c}^{2}+{a}^{2}b{c}^{2}+{a}^{2}{b}^{2}c=\overline{)4}.$

#### Page No 5.27:

#### Question 4:

Factorization of the polynomial $11{x}^{2}-10\sqrt{3}x-3$ gives ____________.

#### Answer:

$11{x}^{2}-10\sqrt{3}x-3\phantom{\rule{0ex}{0ex}}=11{x}^{2}-11\sqrt{3}x+\sqrt{3}x-3\phantom{\rule{0ex}{0ex}}=11x\left(x-\sqrt{3}\right)+\sqrt{3}\left(x-\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(11x+\sqrt{3}\right)\left(x-\sqrt{3}\right)$

Hence, factorization of the polynomial $11{x}^{2}-10\sqrt{3}x-3$ gives $\overline{)\left(11x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}$.

#### Page No 5.27:

#### Question 5:

The polynomial *x*^{2} + *y*^{2} – *z*^{2} – 2*xy* on factorization gives _____________.

#### Answer:

${x}^{2}+{y}^{2}-{z}^{2}-2xy\phantom{\rule{0ex}{0ex}}=\left({x}^{2}+{y}^{2}-2xy\right)-{z}^{2}\phantom{\rule{0ex}{0ex}}={\left(x-y\right)}^{2}-{\left(z\right)}^{2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a-b\right)}^{2}={a}^{2}+{b}^{2}-2ab\right)\phantom{\rule{0ex}{0ex}}=\left(x-y+z\right)\left(x-y-z\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)\right)$

Hence, the polynomial *x*^{2} + *y*^{2} – *z*^{2} – 2*xy* on factorization gives $\overline{)\left(x-y+z\right)\left(x-y-z\right)}$.

#### Page No 5.27:

#### Question 6:

The factors of the expression $a+b+c+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}$ are ____________.

#### Answer:

$a+b+c+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}\phantom{\rule{0ex}{0ex}}={\left(\sqrt{a}\right)}^{2}+{\left(\sqrt{b}\right)}^{2}+{\left(-\sqrt{c}\right)}^{2}+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}\phantom{\rule{0ex}{0ex}}={\left(\sqrt{a}\right)}^{2}+{\left(\sqrt{b}\right)}^{2}+{\left(-\sqrt{c}\right)}^{2}+2\sqrt{a}\sqrt{b}+2\sqrt{b}\left(-\sqrt{c}\right)+2\sqrt{a}\left(-\sqrt{c}\right)\phantom{\rule{0ex}{0ex}}={\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)}^{2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ac={\left(a+b+c\right)}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)$

Hence, the factors of the expression $a+b+c+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}$ are $\overline{)\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)\mathrm{and}\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)}.$.

#### Page No 5.27:

#### Question 7:

The polynomial ,* x*^{6} + 64*y*^{6} on factorization gives _____________.

#### Answer:

${x}^{6}+64{y}^{6}\phantom{\rule{0ex}{0ex}}={\left({x}^{2}\right)}^{3}+{\left(4{y}^{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left({x}^{2}+4{y}^{2}\right)\left({\left({x}^{2}\right)}^{2}+{\left(4{y}^{2}\right)}^{2}-\left({x}^{2}\right)\left(4{y}^{2}\right)\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)\right)\phantom{\rule{0ex}{0ex}}=\left({x}^{2}+4{y}^{2}\right)\left({x}^{4}+16{y}^{4}-4{x}^{2}{y}^{2}\right)$

Hence, the polynomial *x*^{6} + 64*y*^{6} on factorization gives $\overline{)\left({x}^{2}+4{y}^{2}\right)\left({x}^{4}+16{y}^{4}-4{x}^{2}{y}^{2}\right)}$.

#### Page No 5.27:

#### Question 8:

The factorization form of *a*^{4} + *b*^{4 }– *a*^{2}*b*^{2} is _____________.

#### Answer:

${a}^{4}+{b}^{4}-{a}^{2}{b}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{and}\mathrm{subtracting}2{a}^{2}{b}^{2},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}={\left({a}^{2}\right)}^{2}+{\left({b}^{2}\right)}^{2}+2{a}^{2}{b}^{2}-2{a}^{2}{b}^{2}-{a}^{2}{b}^{2}\phantom{\rule{0ex}{0ex}}={\left({a}^{2}+{b}^{2}\right)}^{2}-3{a}^{2}{b}^{2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{a}^{2}+{b}^{2}+2ab={\left(a+b\right)}^{2}\right)\phantom{\rule{0ex}{0ex}}={\left({a}^{2}+{b}^{2}\right)}^{2}-{\left(\sqrt{3}ab\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left({a}^{2}+{b}^{2}+\sqrt{3}ab\right)\left({a}^{2}+{b}^{2}-\sqrt{3}ab\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)\right)$

Hence, the factorization form of *a*^{4} + *b*^{4 }– *a*^{2}*b*^{2} is $\overline{)\left({a}^{2}+{b}^{2}+\sqrt{3}ab\right)\left({a}^{2}+{b}^{2}-\sqrt{3}ab\right)}$.

#### Page No 5.27:

#### Question 9:

If $3x-\frac{y}{5}=10\mathrm{and}xy=5$, then the value of $27{x}^{3}-\frac{{y}^{3}}{125}$ is ____________.

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}3x-\frac{y}{5}=10...\left(1\right)\phantom{\rule{0ex}{0ex}}xy=5...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}3x-\frac{y}{5}=10\phantom{\rule{0ex}{0ex}}\mathrm{Taking}\mathrm{cube}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(3x-\frac{y}{5}\right)}^{3}={10}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(3x\right)}^{3}-{\left(\frac{y}{5}\right)}^{3}-3\left(3x\right)\left(\frac{y}{5}\right)\left(3x-\frac{y}{5}\right)=1000\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a-b\right)}^{3}={a}^{3}-{b}^{3}-3ab\left(a-b\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 27{x}^{3}-\frac{{y}^{3}}{125}-\frac{9}{5}xy\left(3x-\frac{y}{5}\right)=1000\phantom{\rule{0ex}{0ex}}\Rightarrow 27{x}^{3}-\frac{{y}^{3}}{125}-\frac{9}{5}\times 5\times 10=1000\left(\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 27{x}^{3}-\frac{{y}^{3}}{125}-90=1000\phantom{\rule{0ex}{0ex}}\Rightarrow 27{x}^{3}-\frac{{y}^{3}}{125}=1000+90\phantom{\rule{0ex}{0ex}}\Rightarrow 27{x}^{3}-\frac{{y}^{3}}{125}=1090$

Hence, the value of $27{x}^{3}-\frac{{y}^{3}}{125}$ is __1090__.

#### Page No 5.27:

#### Question 10:

The factorized form of $\frac{1}{xyz}\left({x}^{2}+{y}^{2}+{z}^{2}\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is _____________.

#### Answer:

$\frac{1}{xyz}\left({x}^{2}+{y}^{2}+{z}^{2}\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}{xyz}+2\left(\frac{yz+xz+xy}{xyx}\right)\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}+{y}^{2}+{z}^{2}+2\left(yz+xz+xy\right)}{xyz}\phantom{\rule{0ex}{0ex}}=\frac{{\left(x+y+z\right)}^{2}}{xyz}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ac\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{xyz}{\left(x+y+z\right)}^{2}$

Hence, the factorized form of $\frac{1}{xyz}\left({x}^{2}+{y}^{2}+{z}^{2}\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is $\overline{)\frac{1}{xyz}{\left(x+y+z\right)}^{2}}.$

#### Page No 5.27:

#### Question 11:

The factorized form of *a*^{3} + *b*^{3} + 3*ab* – 1 is ____________.

#### Answer:

${a}^{3}+{b}^{3}+3ab-1\phantom{\rule{0ex}{0ex}}={a}^{3}+{b}^{3}-1+3ab\phantom{\rule{0ex}{0ex}}={a}^{3}+{b}^{3}+{\left(-1\right)}^{3}-3ab\left(-1\right)\phantom{\rule{0ex}{0ex}}=\left(a+b-1\right)\left({a}^{2}+{b}^{2}+{\left(-1\right)}^{2}-ab-b\left(-1\right)-a\left(-1\right)\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{a}^{3}+{b}^{3}+{c}^{3}-3abc=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ac\right)\right)\phantom{\rule{0ex}{0ex}}=\left(a+b-1\right)\left({a}^{2}+{b}^{2}+1-ab+b+a\right)$

Hence, the factorized form of *a*^{3} + *b*^{3} + 3*ab* – 1 is $\overline{)\left(a+b-1\right)\left({a}^{2}+{b}^{2}+1-ab+b+a\right)}.$

#### Page No 5.27:

#### Question 1:

If *a* + *b* + *c* = 0, then write the value of *a*^{3} + *b*^{3} + *c*^{3}.

#### Answer:

Recall the formula

When, we have

#### Page No 5.27:

#### Question 2:

If *a*^{2} + *b*^{2} + *c*^{2} = 20 and *a* +* b* +* c* = 0, find* ab* +* bc* + *ca*.

#### Answer:

Recall the formula

Given that

Then we have

#### Page No 5.27:

#### Question 3:

If *a* + *b* + *c* = 9 and *ab* + *bc* + *ca* = 40, find *a*^{2} +* **b*^{2} +*c*^{2}.

#### Answer:

Recall the formula

Given that

,

Then we have

#### Page No 5.27:

#### Question 4:

If *a*^{2} + *b*^{2} + *c*^{2} = 250 and *ab* + *bc* +* ca* = 3, find *a* + *b *+ *c*.

#### Answer:

Recall the formula

Given that

,

Then we have

#### Page No 5.27:

#### Question 5:

Write the value of 25^{3} − 75^{3 }+ 50^{3}.

#### Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

#### Page No 5.27:

#### Question 6:

Write the value of 48^{3} − 30^{3} − 18^{3}.

#### Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

#### Page No 5.28:

#### Question 7:

Write the value of ${\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{3}\right)}^{3}-{\left(\frac{5}{6}\right)}^{3}.$

#### Answer:

The given expression is

Let, and. Then the given expression becomes

Note that:

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

#### Page No 5.28:

#### Question 8:

Write the value of 30^{3} + 20^{3} − 50^{3}.

#### Answer:

The given expression is

Let, and. Then the given expression becomes

Note that

Recall the formula

When, this becomes

So, we have the new formula

, when.

Using the above formula, the value of the given expression is

#### Page No 5.28:

#### Question 9:

Factorize: x^{4} + x^{2} + 25.

#### Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization is.

#### Page No 5.28:

#### Question 10:

Factorize : *x*^{2} − 1 − 2*a* − *a*^{2}

#### Answer:

The given expression to be factorized is

Take commonfrom the last three terms and then we have

We cannot further factorize the expression.

So, the required factorization is.

#### Page No 5.9:

#### Question 1:

*Factorize*:

1. ${x}^{3}+x-3{x}^{2}-3$

#### Answer:

The given expression to be factorized is

Take common *x* from the first two terms and -3 from the last two terms. That is

Finally, take common *x*^{2} + 1from the two terms. That is

We cannot further factorize the expression.

So, the required factorization is.

#### Page No 5.9:

#### Question 2:

*Factorize*:

2. a(a+b)^{3} − 3a^{2}b (a + b)

#### Answer:

The given expression to be factorized is

Take common from the two terms. That is

Expand the term within the second braces.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 3:

$x\left({x}^{3}-{y}^{3}\right)+3xy(x-y)$

#### Answer:

The given expression to be factorized is

We know that

The given expression then becomes

Take common from the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 4:

*Factorize*:

a^{2}x^{2} + (ax^{2} + 1)x + a

#### Answer:

The given expression to be factorized is

Simplify the middle term. That is

Take common from the first two terms and 1 from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 5:

Factorize:

*x ^{2} + y − xy − x*

#### Answer:

The given expression to be factorized is

Rearrange the given expression as

Take common *x* from the first two terms and -1 from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 6:

*Factorize*:

x^{3} − 2x^{2}y + 3xy^{2} − 6y^{3}

#### Answer:

The given expression to be factorized is

Take common from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 7:

*Factorize*:

6*ab* − b^{2} + 12*ac* − 2*bc*

#### Answer:

The given expression to be factorized is

Take common *b* from the first two terms and from the last two terms. That is

Finally, take commonfrom the two terms. That is

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 8:

*Factorize*:

$x(x-2)(x-4)+4x-8$

#### Answer:

The given expression to be factorized is

Take common 4 from the last two terms. That is

Again take commonfrom the two terms of the above expression.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 9:

*Factorize*:

(a − b + c)^{2} + (b − c + a)^{2} + 2(a − b + c) (b − c + a)

#### Answer:

The given expression to be factorized is

This can be written as

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 10:

Factorize:

*a*^{2} + 2*ab** *+*b*^{2} − *c*^{2}

#### Answer:

The given expression to be factorized is

This can be arrange in the form

Substitutingin the above expression, we get.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 11:

Factorize:

a^{2} + 4*b*^{2} − 4*ab* − 4*c*^{2}

#### Answer:

The given expression to be factorized is

This can be arrange in the form

Substitute.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 12:

*Factorize:
x*

^{2}−

*y*

^{2}− 4

*xz*+ 4

*z*

^{2}

#### Answer:

The given expression to be factorized is

Rearrange the terms as

Substitutingin the avove expression,

Put.

We cannot further factorize the expression.

So, the required factorization ofis.

#### Page No 5.9:

#### Question 13:

*Factorize:*

$2{x}^{2}-\frac{5}{6}x+\frac{1}{12}$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 14:

*Factorize:*

${x}^{2}+\frac{12}{35}x+\frac{1}{35}$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 15:

*Factorize:*

$21{x}^{2}-2x+\frac{1}{21}$

#### Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 16:

Give possible expressions for the length and breadth of the rectangle having 35*y*^{2} + 13*y* − 12 as its area.

#### Answer:

The area of the rectangle is

First we will factorize the above expression. This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

The area of a rectangle having length *a* and breadth *b*is *ab*.

Here we don’t know the bigger or the smaller factor. So, the two possibilities are

(i) Length isand breadth is

(ii) Length is and breadth is

#### Page No 5.9:

#### Question 17:

What are the possible expressions for the dimensions of the cuboid whose volume is 3*x*^{2}− 12*x*.

#### Answer:

The volume of the cuboid is

First we will factorize the above expression.

Take commonfrom the two terms of the above expression,

The volume of a cuboid having length, breadth *b* and height is.

Here the word *‘dimensions’* stands for the length, breadth and height of the cuboid. So, the three possibilities are

(i) Length is, breadth is *x* and height is

(ii) Length is *x* , breadth isand height is

(iii) Length is, breadth isand height is *x*

There are many other possibilities also, because we can consider the product of two simple factors as a single factor.

#### Page No 5.9:

#### Question 18:

*Factorize*:

$\left({x}^{2}+\frac{1}{{x}^{2}}\right)-4\left(x+\frac{1}{x}\right)+6$

#### Answer:

The given expression to be factorized is

We have

Use the above result in the original expression to get

Substituting in the above , we get

Put.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 19:

*Factorize*:

(x+2) (x^{2}+25) − 10x^{2} − 20x

#### Answer:

The given expression to be factorized is

Take common from the last two terms. That is

Again take commonfrom the two terms of the above expression. Then

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 20:

*Factorize*:

$2{a}^{2}+2\sqrt{6}ab+3{b}^{2}$

#### Answer:

The given expression to be factorized is

This can be written in the form

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 21:

*Factorize*:

*a*^{2} + *b*^{2} + 2(*ab + bc + ca*)

#### Answer:

The given expression to be factorized is

This can be written as

Take commonfrom the last two terms.

Finally, take common from the two terms of the above expression.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 22:

*Factorize:*

4(x − y)^{2} − 12(x − y) (x + y) + 9(x + y)^{2}

#### Answer:

The given expression to be factorized is

Substituting andin the above expression, we get

=

This can be arrange in the form

Putand.

Take common -1 from the expression within the braces.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 23:

*Factorize:*

a^{2} + b^{2} + 2bc − c^{2}

#### Answer:

The given expression to be factorized is

This can be arrange in the form

Substitutingin the above expression, we get.

Put.

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 24:

*Factorize:
xy*

^{9}−

*yx*

^{9}

#### Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the two terms of the above expression

We cannot further factorize the expression.

So, the required factorization of is

#### Page No 5.9:

#### Question 25:

*Factorize:*

*x*^{4}^{ }+ *x*^{2}*y*^{2} + *y*^{4}

#### Answer:

The given expression to be factorized is

Add and subtract the termin the given expression.

Substitutingin the above expression, we get

Putin the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 26:

*Factorize:
${x}^{2}+6\sqrt{2}x+10$*

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms.

Finally, take commonfrom the above expression. Then we have

We cannot further factorize the expression.

So, the required factorization is.

#### Page No 5.9:

#### Question 27:

*Factorize:*

${x}^{2}-2\sqrt{2}x-30$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 28:

*Factorize:*

${x}^{2}-\sqrt{3}x-6$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms. Then we have

Finally take commonfrom the above expression. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 29:

*Factorize:*

${x}^{2}+5\sqrt{5}x+30$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms. Then we have

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 30:

*Factorize:*

${x}^{2}+2\sqrt{3}x-24$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take common *x* from the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 31:

*Factorize:*

$5\sqrt{5}{x}^{2}+20x+3\sqrt{5}$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 32:

*Factorize:*

$2{x}^{2}+3\sqrt{5}x+5$

#### Answer:

The given expression to be factorized is

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 33:

*Factorize:*

9(2*a* − *b*)^{2} − 4(2*a* − *b*) − 13

#### Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take common *x* from the first two terms and 1 from the last two terms,

Finally take commonfrom the above expression,

Put,

We cannot further factorize the expression.

So, the required factorization ofis.

#### Page No 5.9:

#### Question 34:

*Factorize:*

7(*x* − 2*y*)^{2} − 25(*x* − 2*y*) + 12

#### Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

Put in the above expression,

We cannot further factorize the expression.

So, the required factorization of is.

#### Page No 5.9:

#### Question 35:

*Factorize:*

2(*x* + *y*)^{2} − 9(*x* + *y*) − 5

#### Answer:

The given expression to be factorized is

Substitutingin the above expression, we get

This can be written in the form

Take commonfrom the first two terms andfrom the last two terms,

Finally take commonfrom the above expression,

Put. Then we have

We cannot further factorize the expression.

So, the required factorization of is.

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