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#### Page No 99:

We have:
$9{x}^{2}+12xy\phantom{\rule{0ex}{0ex}}=3x\left(3x+4y\right)$

#### Page No 99:

We have:
$18{x}^{2}y-24xyz\phantom{\rule{0ex}{0ex}}=6xy\left(3y-4z\right)$

#### Page No 99:

We have:
$27{a}^{3}{b}^{3}-45{a}^{4}{b}^{2}\phantom{\rule{0ex}{0ex}}=9{a}^{3}{b}^{2}\left(3b-5a\right)$

#### Page No 99:

We have:
$2a\left(x+y\right)-3b\left(x+y\right)\phantom{\rule{0ex}{0ex}}=\left(x+y\right)\left(2a-3b\right)$

#### Page No 99:

We have:
$2x\left({p}^{2}+{q}^{2}\right)+4y\left({p}^{2}+{q}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\left[x\left({p}^{2}+{q}^{2}\right)+2y\left({p}^{2}+{q}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\left({p}^{2}+{q}^{2}\right)\left(x+2y\right)\phantom{\rule{0ex}{0ex}}$

#### Page No 99:

We have:
$x\left(a-5\right)+y\left(5-a\right)=x\left(a-5\right)-y\left(a-5\right)$
$=\left(a-5\right)\left(x-y\right)$

#### Page No 99:

We have:
$4\left(a+b\right)-6{\left(a+b\right)}^{2}=2\left(a+b\right)\left[2-3\left(a+b\right)\right]$
$=2\left(a+b\right)\left(2-3a-3b\right)$

#### Page No 99:

We have:
$8{\left(3a-2b\right)}^{2}-10\left(3a-2b\right)=2\left(3a-2b\right)\left[4\left(3a-2b\right)-5\right]$
$=2\left(3a-2b\right)\left(12a-8b-5\right)$

#### Page No 99:

We have:
$x{\left(x+y\right)}^{3}-3{x}^{2}y\left(x+y\right)=x\left(x+y\right)\left[{\left(x+y\right)}^{2}-3xy\right]\phantom{\rule{0ex}{0ex}}$
$=x\left(x+y\right)\left[{x}^{2}+{y}^{2}+2xy-3xy\right]\phantom{\rule{0ex}{0ex}}=x\left(x+y\right)\left({x}^{2}+{y}^{2}-xy\right)$

#### Page No 99:

We have:
${x}^{3}+2{x}^{2}+5x+10=\left({x}^{3}+2{x}^{2}\right)+\left(5x+10\right)$
$={x}^{2}\left(x+2\right)+5\left(x+2\right)\phantom{\rule{0ex}{0ex}}=\left(x+2\right)\left({x}^{2}+5\right)$

We have:

#### Page No 99:

We have:

$=b\left[{a}^{2}\left(a-1\right)+5\left(a-1\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=b\left(a-1\right)\left({a}^{2}+5\right)$

#### Page No 99:

We have:

#### Page No 99:

We have:
${x}^{3}-2{x}^{2}y+3x{y}^{2}-6{y}^{3}=\left({x}^{3}-2{x}^{2}y\right)+\left(3x{y}^{2}-6{y}^{3}\right)$
$={x}^{2}\left(x-2y\right)+3{y}^{2}\left(x-2y\right)\phantom{\rule{0ex}{0ex}}=\left(x-2y\right)\left({x}^{2}+3{y}^{2}\right)$

#### Page No 99:

We have:
$px-5q+pq-5x=\left(px-5x\right)+\left(pq-5q\right)$
$=x\left(p-5\right)+q\left(p-5\right)\phantom{\rule{0ex}{0ex}}=\left(p-5\right)\left(x+q\right)$

#### Page No 99:

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

#### Page No 99:

We have:
${\left(3a-1\right)}^{2}-6a+2={\left(3a-1\right)}^{2}-2\left(3a-1\right)$
$=\left(3a-1\right)\left[\left(3a-1\right)-2\right]\phantom{\rule{0ex}{0ex}}=\left(3a-1\right)\left(3a-1-2\right)\phantom{\rule{0ex}{0ex}}=\left(3a-1\right)\left(3a-3\right)\phantom{\rule{0ex}{0ex}}=3\left(3a-1\right)\left(a-1\right)$

#### Page No 99:

We have:
${\left(2x-3\right)}^{2}-8x+12={\left(2x-3\right)}^{2}-4\left(2x-3\right)$
$=\left(2x-3\right)\left[\left(2x-3\right)-4\right]\phantom{\rule{0ex}{0ex}}=\left(2x-3\right)\left(2x-3-4\right)\phantom{\rule{0ex}{0ex}}=\left(2x-3\right)\left(2x-7\right)$

#### Page No 99:

We have:
${a}^{3}+a-3{a}^{2}-3=\left({a}^{3}-3{a}^{2}\right)+\left(a-3\right)\phantom{\rule{0ex}{0ex}}$
$={a}^{2}\left(a-3\right)+1\left(a-3\right)\phantom{\rule{0ex}{0ex}}=\left(a-3\right)\left({a}^{2}+1\right)$

#### Page No 99:

We have:
$3ax-6ay-8by+4bx=\left(3ax-6ay\right)+\left(4bx-8by\right)$
$=3a\left(x-2y\right)+4b\left(x-2y\right)\phantom{\rule{0ex}{0ex}}=\left(x-2y\right)\left(3a+4b\right)$

#### Page No 99:

We have:
$ab{x}^{2}+{a}^{2}x+{b}^{2}x+ab=\left(ab{x}^{2}+{b}^{2}x\right)+\left({a}^{2}x+ab\right)$
$=bx\left(ax+b\right)+a\left(ax+b\right)\phantom{\rule{0ex}{0ex}}=\left(ax+b\right)\left(bx+a\right)$

#### Page No 99:

We have:
${x}^{3}-{x}^{2}+ax+x-a-1=\left({x}^{3}-{x}^{2}\right)+\left(ax-a\right)+\left(x-1\right)$
$={x}^{2}\left(x-1\right)+a\left(x-1\right)+1\left(x-1\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left({x}^{2}+a+1\right)$

#### Page No 100:

We have:
$2x+4y-8xy-1=\left(2x-8xy\right)-\left(1-4y\right)$
$=2x\left(1-4y\right)-1\left(1-4y\right)\phantom{\rule{0ex}{0ex}}=\left(1-4y\right)\left(2x-1\right)$

#### Page No 100:

We have:
$ab\left({x}^{2}+{y}^{2}\right)-xy\left({a}^{2}+{b}^{2}\right)=ab{x}^{2}+ab{y}^{2}-{a}^{2}xy-{b}^{2}xy$
$=\left(ab{x}^{2}-{a}^{2}xy\right)-\left({b}^{2}xy-ab{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=ax\left(bx-ay\right)-by\left(bx-ay\right)\phantom{\rule{0ex}{0ex}}=\left(bx-ay\right)\left(ax-by\right)$

#### Page No 100:

We have:
${a}^{2}+ab\left(b+1\right)+{b}^{3}={a}^{2}+a{b}^{2}+ab+{b}^{3}$
$=\left({a}^{2}+a{b}^{2}\right)+\left(ab+{b}^{3}\right)\phantom{\rule{0ex}{0ex}}=a\left(a+{b}^{2}\right)+b\left(a+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(a+{b}^{2}\right)\left(a+b\right)$

#### Page No 100:

We have:
${a}^{3}+ab\left(1-2a\right)-2{b}^{2}={a}^{3}+ab-2{a}^{2}b-2{b}^{2}$
$=\left({a}^{3}-2{a}^{2}b\right)+\left(ab-2{b}^{2}\right)\phantom{\rule{0ex}{0ex}}={a}^{2}\left(a-2b\right)+b\left(a-2b\right)\phantom{\rule{0ex}{0ex}}=\left(a-2b\right)\left({a}^{2}+b\right)$

#### Page No 100:

We have:
$2{a}^{2}+bc-2ab-ac=\left(2{a}^{2}-2ab\right)-\left(ac-bc\right)$
$=2a\left(a-b\right)-c\left(a-b\right)\phantom{\rule{0ex}{0ex}}=\left(a-b\right)\left(2a-c\right)$

#### Page No 100:

We have:
${\left(ax+by\right)}^{2}+{\left(bx-ay\right)}^{2}=\left[{\left(ax\right)}^{2}+2×ax×by+{\left(by\right)}^{2}\right]+\left[{\left(bx\right)}^{2}-2×bx×ay+{\left(ay\right)}^{2}\right]$
$={a}^{2}{x}^{2}+2abxy+{b}^{2}{y}^{2}+{b}^{2}{x}^{2}-2abxy+{a}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}{x}^{2}+{b}^{2}{y}^{2}+{b}^{2}{x}^{2}+{a}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}=\left({a}^{2}{x}^{2}+{b}^{2}{x}^{2}\right)+\left({a}^{2}{y}^{2}+{b}^{2}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({a}^{2}+{b}^{2}\right)+{y}^{2}\left({a}^{2}+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{2}+{b}^{2}\right)\left({x}^{2}+{y}^{2}\right)$

#### Page No 100:

We have:
$a\left(a+b-c\right)-bc={a}^{2}+ab-ac-bc$
$=\left({a}^{2}-ac\right)+\left(ab-bc\right)\phantom{\rule{0ex}{0ex}}=a\left(a-c\right)+b\left(a-c\right)\phantom{\rule{0ex}{0ex}}=\left(a-c\right)\left(a+b\right)$

#### Page No 100:

We have:
$a\left(a-2b-c\right)+2bc={a}^{2}-2ab-ac+2bc$
$=\left({a}^{2}-2ab\right)-\left(ac-2bc\right)\phantom{\rule{0ex}{0ex}}=a\left(a-2b\right)-c\left(a-2b\right)\phantom{\rule{0ex}{0ex}}=\left(a-2b\right)\left(a-c\right)$

#### Page No 100:

We have:
${a}^{2}{x}^{2}+\left(a{x}^{2}+1\right)x+a=\left(a{x}^{2}+1\right)x+\left({a}^{2}{x}^{2}+a\right)$
$=x\left(a{x}^{2}+1\right)+a\left(a{x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}=\left(a{x}^{2}+1\right)\left(x+a\right)$

#### Page No 100:

We have:
$ab\left({x}^{2}+1\right)+x\left({a}^{2}+{b}^{2}\right)=ab{x}^{2}+ab+{a}^{2}x+{b}^{2}x$
$=\left(ab{x}^{2}+{a}^{2}x\right)+\left({b}^{2}x+ab\right)\phantom{\rule{0ex}{0ex}}=ax\left(bx+a\right)+b\left(bx+a\right)\phantom{\rule{0ex}{0ex}}=\left(bx+a\right)\left(ax+b\right)$

#### Page No 100:

We have:
${x}^{2}-\left(a+b\right)x+ab={x}^{2}-ax-bx+ab$
$=\left({x}^{2}-ax\right)-\left(bx-ab\right)\phantom{\rule{0ex}{0ex}}=x\left(x-a\right)-b\left(x-a\right)\phantom{\rule{0ex}{0ex}}=\left(x-a\right)\left(x-b\right)$

#### Page No 105:

$1+2ab-\left({a}^{2}+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=1+2ab-{a}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}=1-{a}^{2}+2ab-{b}^{2}\phantom{\rule{0ex}{0ex}}={1}^{2}-\left({a}^{2}-2ab+{b}^{2}\right)$

#### Page No 105:

Disclaimer: The expression of the question should be ${x}^{2}-2+\frac{1}{{x}^{2}}-{y}^{2}$. The same has been done before solving the question.

#### Page No 114:

We have:
${x}^{2}+11x+30$
We have to split 11 into two numbers such that their sum of is 11 and their product is 30.
Clearly, .

#### Page No 114:

We have:
${x}^{2}+18x+32$
We have to split 18 into two numbers such that their sum is 18 and their product is 32.
Clearly, .

#### Page No 114:

${x}^{2}+20x-69\phantom{\rule{0ex}{0ex}}={x}^{2}+23x-3x-69\phantom{\rule{0ex}{0ex}}=x\left(x+23\right)-3\left(x+23\right)\phantom{\rule{0ex}{0ex}}=\left(x+23\right)\left(x-3\right)$

#### Page No 114:

${x}^{2}+19x-150\phantom{\rule{0ex}{0ex}}={x}^{2}+25x-6x-150\phantom{\rule{0ex}{0ex}}=x\left(x+25\right)-6\left(x+25\right)\phantom{\rule{0ex}{0ex}}=\left(x+25\right)\left(x-6\right)$

#### Page No 114:

${x}^{2}+7x-98\phantom{\rule{0ex}{0ex}}={x}^{2}+14x-7x-98\phantom{\rule{0ex}{0ex}}=x\left(x+14\right)-7\left(x+14\right)\phantom{\rule{0ex}{0ex}}=\left(x+14\right)\left(x-7\right)$

#### Page No 114:

${x}^{2}+2\sqrt{3}x–24\phantom{\rule{0ex}{0ex}}={x}^{2}+4\sqrt{3}x-2\sqrt{3}x-24\phantom{\rule{0ex}{0ex}}=x\left(x+4\sqrt{3}\right)-2\sqrt{3}\left(x+4\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x+4\sqrt{3}\right)\left(x-2\sqrt{3}\right)$

#### Page No 114:

${x}^{2}-21x+90\phantom{\rule{0ex}{0ex}}={x}^{2}-15x-6x+90\phantom{\rule{0ex}{0ex}}=x\left(x-15\right)-6\left(x-15\right)\phantom{\rule{0ex}{0ex}}=\left(x-6\right)\left(x-15\right)$

#### Page No 114:

${x}^{2}-22x+120\phantom{\rule{0ex}{0ex}}={x}^{2}-12x-10x+120\phantom{\rule{0ex}{0ex}}=x\left(x-12\right)-10\left(x-12\right)\phantom{\rule{0ex}{0ex}}=\left(x-10\right)\left(x-12\right)$

#### Page No 114:

${x}^{2}-4x+3\phantom{\rule{0ex}{0ex}}={x}^{2}-3x-x+3\phantom{\rule{0ex}{0ex}}=x\left(x-3\right)-1\left(x-3\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-3\right)$

#### Page No 114:

${x}^{2}+7\sqrt{6}x+60\phantom{\rule{0ex}{0ex}}={x}^{2}+5\sqrt{6}x+2\sqrt{6}x+60\phantom{\rule{0ex}{0ex}}=x\left(x+5\sqrt{6}\right)+2\sqrt{6}\left(x+5\sqrt{6}\right)\phantom{\rule{0ex}{0ex}}=\left(x+5\sqrt{6}\right)\left(x+2\sqrt{6}\right)$

#### Page No 114:

${x}^{2}+3\sqrt{3}x+6\phantom{\rule{0ex}{0ex}}={x}^{2}+2\sqrt{3}x+\sqrt{3}x+6\phantom{\rule{0ex}{0ex}}=x\left(x+2\sqrt{3}\right)+\sqrt{3}\left(x+2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x+2\sqrt{3}\right)\left(x+\sqrt{3}\right)$

#### Page No 114:

${x}^{2}+6\sqrt{6}x+48\phantom{\rule{0ex}{0ex}}={x}^{2}+4\sqrt{6}x+2\sqrt{6}x+48\phantom{\rule{0ex}{0ex}}=x\left(x+4\sqrt{6}\right)+2\sqrt{6}\left(x+4\sqrt{6}\right)\phantom{\rule{0ex}{0ex}}=\left(x+4\sqrt{6}\right)\left(x+2\sqrt{6}\right)$

#### Page No 114:

${x}^{2}+5\sqrt{5}x+30\phantom{\rule{0ex}{0ex}}={x}^{2}+3\sqrt{5}x+2\sqrt{5}x+30\phantom{\rule{0ex}{0ex}}=x\left(x+3\sqrt{5}\right)+2\sqrt{5}\left(x+3\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\left(x+3\sqrt{5}\right)\left(x+2\sqrt{5}\right)$

#### Page No 114:

${x}^{2}-24x-180\phantom{\rule{0ex}{0ex}}={x}^{2}-30x+6x-180\phantom{\rule{0ex}{0ex}}=x\left(x-30\right)+6\left(x-30\right)\phantom{\rule{0ex}{0ex}}=\left(x-30\right)\left(x+6\right)$

#### Page No 114:

${x}^{2}-32x-105\phantom{\rule{0ex}{0ex}}={x}^{2}-35x+3x-105\phantom{\rule{0ex}{0ex}}=x\left(x-35\right)+3\left(x-35\right)\phantom{\rule{0ex}{0ex}}=\left(x-35\right)\left(x+3\right)$

#### Page No 114:

${x}^{2}-11x-80\phantom{\rule{0ex}{0ex}}={x}^{2}-16x+5x-80\phantom{\rule{0ex}{0ex}}=x\left(x-16\right)+5\left(x-16\right)\phantom{\rule{0ex}{0ex}}=\left(x-16\right)\left(x+5\right)$

#### Page No 114:

$-{x}^{2}-x+6\phantom{\rule{0ex}{0ex}}=-{x}^{2}-3x+2x+6\phantom{\rule{0ex}{0ex}}=-x\left(x+3\right)+2\left(x+3\right)\phantom{\rule{0ex}{0ex}}=\left(x+3\right)\left(-x+2\right)\phantom{\rule{0ex}{0ex}}=\left(x+3\right)\left(2-x\right)$

#### Page No 114:

${x}^{2}-\sqrt{3}x-6\phantom{\rule{0ex}{0ex}}={x}^{2}-2\sqrt{3}x+\sqrt{3}x-6\phantom{\rule{0ex}{0ex}}=x\left(x-2\sqrt{3}\right)+\sqrt{3}\left(x-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x-2\sqrt{3}\right)\left(x+\sqrt{3}\right)$

#### Page No 114:

$-{x}^{2}+3x+40\phantom{\rule{0ex}{0ex}}=-{x}^{2}+8x-5x+40\phantom{\rule{0ex}{0ex}}=-x\left(x-8\right)-5\left(x-8\right)\phantom{\rule{0ex}{0ex}}=\left(x-8\right)\left(-x-5\right)\phantom{\rule{0ex}{0ex}}=\left(8-x\right)\left(x+5\right)$

#### Page No 114:

${x}^{2}-26x+133\phantom{\rule{0ex}{0ex}}={x}^{2}-19x-7x+133\phantom{\rule{0ex}{0ex}}=x\left(x-19\right)-7\left(x-19\right)\phantom{\rule{0ex}{0ex}}=\left(x-19\right)\left(x-7\right)$

#### Page No 114:

${x}^{2}-2\sqrt{3}x-24\phantom{\rule{0ex}{0ex}}={x}^{2}-4\sqrt{3}x+2\sqrt{3}x-24\phantom{\rule{0ex}{0ex}}=x\left(x-4\sqrt{3}\right)+2\sqrt{3}\left(x-4\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x-4\sqrt{3}\right)\left(x+2\sqrt{3}\right)$

#### Page No 114:

${x}^{2}-3\sqrt{5}x-20\phantom{\rule{0ex}{0ex}}={x}^{2}-4\sqrt{5}x+\sqrt{5}x-20\phantom{\rule{0ex}{0ex}}=x\left(x-4\sqrt{5}\right)+\sqrt{5}\left(x-4\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\left(x-4\sqrt{5}\right)\left(x+\sqrt{5}\right)$

#### Page No 114:

${x}^{2}+\sqrt{2}x-24\phantom{\rule{0ex}{0ex}}={x}^{2}+4\sqrt{2}x-3\sqrt{2}x-24\phantom{\rule{0ex}{0ex}}=x\left(x+4\sqrt{2}\right)-3\sqrt{2}\left(x+4\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}=\left(x+4\sqrt{2}\right)\left(x-3\sqrt{2}\right)$

#### Page No 114:

${x}^{2}-2\sqrt{2}x-30\phantom{\rule{0ex}{0ex}}={x}^{2}-5\sqrt{2}x+3\sqrt{2}x-30\phantom{\rule{0ex}{0ex}}=x\left(x-5\sqrt{2}\right)+3\sqrt{2}\left(x-5\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}=\left(x-5\sqrt{2}\right)\left(x+3\sqrt{2}\right)$

#### Page No 114:

We have:
${x}^{2}-x-156$
We have to split ($-$1) into two numbers such that their sum is ($-$1) and their product is ($-$156).
Clearly, .

#### Page No 114:

${x}^{2}-32x-105\phantom{\rule{0ex}{0ex}}={x}^{2}-35x+3x-105\phantom{\rule{0ex}{0ex}}=x\left(x-35\right)+3\left(x-35\right)\phantom{\rule{0ex}{0ex}}=\left(x-35\right)\left(x+3\right)$

#### Page No 114:

$9{x}^{2}+18x+8\phantom{\rule{0ex}{0ex}}=9{x}^{2}+12x+6x+8\phantom{\rule{0ex}{0ex}}=3x\left(3x+4\right)+2\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=\left(3x+4\right)\left(3x+2\right)$

#### Page No 114:

$6{x}^{2}+17x+12\phantom{\rule{0ex}{0ex}}=6{x}^{2}+9x+8x+12\phantom{\rule{0ex}{0ex}}=3x\left(2x+3\right)+4\left(2x+3\right)\phantom{\rule{0ex}{0ex}}=\left(2x+3\right)\left(3x+4\right)$

#### Page No 114:

We have:
$18{x}^{2}+3x-10$
We have to split 3 into two numbers such that their sum is 3 and their product is ($-$180), i.e., $18×\left(-10\right)$.
Clearly, .

â€‹

#### Page No 114:

We have:
$2{x}^{2}+11x-21$
We have to split 11 into two numbers such that their sum is 11 and their product is ($-$42), i.e., $2×\left(-21\right)$.
Clearly, .

#### Page No 114:

We have:
$15{x}^{2}+2x-8$
We have to split 2 into two numbers such that their sum is 2 and their product is ($-$120), i.e., $15×\left(-8\right)$.
Clearly, .

#### Page No 114:

$21{x}^{2}+5x-6\phantom{\rule{0ex}{0ex}}=21{x}^{2}+14x-9x-6\phantom{\rule{0ex}{0ex}}=7x\left(3x+2\right)-3\left(3x+2\right)\phantom{\rule{0ex}{0ex}}=\left(3x+2\right)\left(7x-3\right)$

#### Page No 114:

We have:
$24{x}^{2}-41x+12$
We have to split ($-$41) into two numbers such that their sum is ($-$41) and their product is 288, i.e., $24×12$.
Clearly, .

#### Page No 114:

Hence, factorisation of 3x2 – 14x + 8 is $\left(x-4\right)\left(3x-2\right)$.

#### Page No 114:

We have:
$2{x}^{2}+3x-90$
We have to split 3 into two numbers such that their sum is 3 and their product is ($-$180), i.e., $2×\left(-90\right)$.
Clearly, .

#### Page No 114:

We have:
$\sqrt{5}{x}^{2}+2x-3\sqrt{5}$
We have to split 2 into two numbers such that their sum is 2 and product is ($-$15), i.e.,$\sqrt{5}×\left(-3\sqrt{5}\right)$.
Clearly, .

#### Page No 114:

We have:
$2\sqrt{3}{x}^{2}+x-5\sqrt{3}$
We have to split 1 into two numbers such that their sum is 1 and product is 30, i.e.,$2\sqrt{3}×\left(-5\sqrt{3}\right)$.
Clearly, .

#### Page No 114:

We have:
$7{x}^{2}+2\sqrt{14}x+2$
We have to split $2\sqrt{14}$ into two numbers such that their sum is $2\sqrt{14}$ and product is 14.
Clearly, .

#### Page No 114:

We have:
$6\sqrt{3}{x}^{2}-47x+5\sqrt{3}$
Now, we have to split ($-$47) into two numbers such that their sum is ($-$47) and their product is 90.
Clearly, .

#### Page No 114:

We have:
$5\sqrt{5}{x}^{2}+20x+3\sqrt{5}$
We have to split 20 into two numbers such that their sum is 20 and their product is 75.
Clearly,

#### Page No 114:

Hence, factorisation of $\sqrt{3}{x}^{2}+10x+8\sqrt{3}$ is $\left(x+2\sqrt{3}\right)\left(\sqrt{3}x+4\right)$.

#### Page No 114:

We have:
$\sqrt{2}{x}^{2}+3x+\sqrt{2}$
We have to split 3 into two numbers such that their sum is 3 and their product is 2, i.e., $\sqrt{2}×\sqrt{2}$.
Clearly, .

#### Page No 114:

We have:
$2{x}^{2}+3\sqrt{3}x+3$
We have to split $3\sqrt{3}$ into two numbers such that their sum is $3\sqrt{3}$ and their product is 6, i.e.,$2×3$.
Clearly, .

#### Page No 114:

We have:
$15{x}^{2}-x-28$
We have to split ($-$1) into two numbers such that their sum is ($-$1) and their product is ($-$420), i.e., $15×\left(-28\right)$.
Clearly, .

#### Page No 114:

We have:
$6{x}^{2}-5x-21$
We have to split ($-$5) into two numbers such that their sum is ($-$5) and their product is ($-$126), i.e., $6×\left(-21\right)$.
Clearly, .

#### Page No 114:

We have:
$2{x}^{2}-7x-15$
We have to split ($-$7) into two numbers such that their sum is ($-$7) and their product is ($-$30), i.e., $2×\left(-15\right)$.
Clearly, .

#### Page No 114:

We have:
$5{x}^{2}-16x-21$
We have to split ($-$16) into two numbers such that their sum is ($-$16) and their product is ($-$105), i.e., $5×\left(-21\right)$.
Clearly, .

#### Page No 114:

Hence, factorisation of 6x2 – 11x – 35 is $\left(2x-7\right)\left(3x+5\right)$.

#### Page No 114:

Hence, factorisation of 9x2 – 3x – 20 is $\left(3x-5\right)\left(3x+4\right)$.

#### Page No 114:

We have:
$10{x}^{2}-9x-7$

We have to split ($-$9) into two numbers such that their sum is ($-$9) and their product is ($-$70), i.e., $10×\left(-7\right)$.
Clearly, .

#### Page No 114:

Now, we have to split ($-$32) into two numbers such that their sum is ($-$32) and their product is 112, i.e., $16×7$.
Clearly, .

#### Page No 114:

Hence, factorisation of $\frac{1}{3}{x}^{2}-2x-9$ is $\left(\frac{1}{3}x-3\right)\left(x+3\right)$.

#### Page No 114:

Hence, factorisation of ${x}^{2}+\frac{12}{35}x+\frac{1}{35}$ is $\left(x+\frac{1}{5}\right)\left(x+\frac{1}{7}\right)$.

#### Page No 114:

Hence, factorisation of $21{x}^{2}-2x+\frac{1}{21}$ is $\left(x-\frac{1}{21}\right)\left(21x-1\right)$.

#### Page No 114:

Hence, factorisation of $\frac{3}{2}{x}^{2}+16x+10$ is $\left(x+10\right)\left(\frac{3}{2}x+1\right)$.

#### Page No 114:

Hence, factorisation of $\frac{2}{3}{x}^{2}-\frac{17}{3}x-28$ is $\left(\frac{1}{3}x-4\right)\left(2x+7\right)$.

#### Page No 114:

Hence, factorisation of $\frac{3}{5}{x}^{2}-\frac{19}{5}x+4$ is $\left(\frac{1}{5}x-1\right)\left(3x-4\right)$.

#### Page No 114:

Hence, factorisation of $2{x}^{2}-x+\frac{1}{8}$ is $\left(x-\frac{1}{4}\right)\left(2x-\frac{1}{2}\right)$.

#### Page No 114:

We have:
$2{\left(x+y\right)}^{2}-9\left(x+y\right)-5\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\left(x+y\right)=u\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Thus, the given expression becomes
$2{u}^{2}-9u-5\phantom{\rule{0ex}{0ex}}$
Now, we have to split ($-$9) into two numbers such that their sum is ($-$9) and their product is ($-$10).
Clearly, .

Putting $u=\left(x+y\right)$, we get:

#### Page No 114:

We have:
$9\left(2a-b{\right)}^{2}-4\left(2a-b\right)-13\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\left(2a-b\right)=p\phantom{\rule{0ex}{0ex}}$
Thus, the given expression becomes
$9{p}^{2}-4p-13\phantom{\rule{0ex}{0ex}}$
Now, we must split ($-$4) into two numbers such that their sum is ($-$4) and their product is ($-$117).
Clearly, .

Putting $p=\left(2a-b\right)$, we get:

#### Page No 115:

Hence, factorisation of $7{\left(x-2y\right)}^{2}-25\left(x-2y\right)+12$ is $\left(7x-14y-4\right)\left(x-2y-3\right)$.

#### Page No 115:

Hence, factorisation of $10{\left(3x+\frac{1}{x}\right)}^{2}-\left(3x+\frac{1}{x}\right)-3$ is $\left(15x+\frac{5}{x}-3\right)\left(6x+\frac{2}{x}+1\right)$.

#### Page No 115:

Hence, factorisation of $6{\left(2x-\frac{3}{x}\right)}^{2}+7\left(2x-\frac{3}{x}\right)-20$ is $\left(4x-\frac{6}{x}+5\right)\left(6x-\frac{9}{x}-4\right)$.

#### Page No 115:

Hence, factorisation of ${\left(a+2b\right)}^{2}+101\left(a+2b\right)+100$ is $\left(a+2b+1\right)\left(a+2b+100\right)$.

#### Page No 115:

Hence, factorisation of 4x4 + 7x2 – 2 is $\left(4{x}^{2}-1\right)\left({x}^{2}+2\right)$.

#### Page No 115:

Hence, {(999)2 – 1} = 998000.

#### Page No 119:

Hence, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz = ${\left(4x-2y+3z\right)}^{2}$.

#### Page No 123:

Hence, factorisation of $8{a}^{3}+27{b}^{3}+36{a}^{2}b+54a{b}^{2}$ is ${\left(2a+3b\right)}^{3}$.

#### Page No 123:

Hence, factorisation of $64{a}^{3}-27{b}^{3}-144{a}^{2}b+108a{b}^{2}$ is ${\left(4a-3b\right)}^{3}$.

#### Page No 123:

Hence, factorisation of $1+\frac{27}{125}{a}^{3}+\frac{9a}{5}+\frac{27{a}^{2}}{25}$ is ${\left(1+\frac{3}{5}a\right)}^{3}$.

#### Page No 123:

Hence, factorisation of $125{x}^{3}-27{y}^{3}-225{x}^{2}y+135x{y}^{2}$ is ${\left(5x-3y\right)}^{3}$.

#### Page No 123:

Hence, factorisation of ${a}^{3}{x}^{3}-3{a}^{2}b{x}^{2}+3a{b}^{2}x-{b}^{3}$ is ${\left(ax-b\right)}^{3}$.

#### Page No 123:

Hence, factorisation of $\frac{64}{125}{a}^{3}-\frac{96}{25}{a}^{2}+\frac{48}{5}a-8$ is ${\left(\frac{4}{5}a-2\right)}^{3}$.

#### Page No 123:

Hence, factorisation of a3 – 12a(a – 4) – 64 is ${\left(a-4\right)}^{3}$.

#### Page No 129:

We know that
${x}^{3}+{y}^{3}=\left(x+y\right)\left({x}^{2}+{y}^{2}-xy\right)$
Given: 27a3 + 64b3
x = 3a, y = 4b

#### Page No 129:

We know
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$
We have,
$\frac{{x}^{3}}{216}-8{y}^{3}={\left(\frac{x}{6}\right)}^{3}-{\left(2y\right)}^{3}$
So, $a=\frac{x}{6},b=2y$
$\frac{{x}^{3}}{216}-8{y}^{3}=\left(\frac{x}{6}-2y\right)\left({\left(\frac{x}{6}\right)}^{2}+\frac{x}{6}×2y+{\left(2y\right)}^{2}\right)=\left(\frac{x}{6}-2y\right)\left(\frac{{x}^{2}}{36}+\frac{xy}{3}+4{y}^{2}\right)$

#### Page No 129:

Using the identity
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$

#### Page No 129:

$8{x}^{2}{y}^{3}–{x}^{5}={x}^{2}\left(8{y}^{3}-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(2y-x\right)\left(4{y}^{2}+{x}^{2}+2xy\right)$

#### Page No 129:

$1029–3{x}^{3}\phantom{\rule{0ex}{0ex}}=3\left(343-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=3\left({7}^{3}-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=3\left(7-x\right)\left(49+{x}^{2}+7x\right)$

#### Page No 129:

a12 – b12
$=\left({a}^{6}+{b}^{6}\right)\left({a}^{6}-{b}^{6}\right)\phantom{\rule{0ex}{0ex}}=\left[{\left({a}^{2}\right)}^{3}+{\left({b}^{2}\right)}^{3}\right]\left[{\left({a}^{3}\right)}^{2}-{\left({b}^{3}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\left[\left({a}^{2}+{b}^{2}\right)\left({a}^{4}+{b}^{4}-{a}^{2}{b}^{2}\right)\right]\left[\left({a}^{3}-{b}^{3}\right)\left({a}^{3}+{b}^{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\left[\left({a}^{2}+{b}^{2}\right)\left({a}^{4}+{b}^{4}-{a}^{2}{b}^{2}\right)\right]\left[\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)\right]\phantom{\rule{0ex}{0ex}}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)\left({a}^{2}+{b}^{2}\right)\left({a}^{4}+{b}^{4}-{a}^{2}{b}^{2}\right)$

#### Page No 129:

Let ${x}^{3}=y$
So, the equation becomes
${y}^{2}-7y-8={y}^{2}-8y+y-8\phantom{\rule{0ex}{0ex}}=y\left(y-8\right)+\left(y-8\right)\phantom{\rule{0ex}{0ex}}=\left(y-8\right)\left(y+1\right)\phantom{\rule{0ex}{0ex}}=\left({x}^{3}-8\right)\left({x}^{3}+1\right)\phantom{\rule{0ex}{0ex}}=\left(x-2\right)\left({x}^{2}+4+2x\right)\left(x+1\right)\left({x}^{2}+1-x\right)$

#### Page No 129:

x3 – 3x+ 3x + 7
$={x}^{3}–3{x}^{2}+3x+7\phantom{\rule{0ex}{0ex}}={x}^{3}–3{x}^{2}+3x+8-1={x}^{3}–3{x}^{2}+3x-1+8\phantom{\rule{0ex}{0ex}}=\left({x}^{3}–3{x}^{2}+3x-1\right)+8\phantom{\rule{0ex}{0ex}}={\left(x-1\right)}^{3}+{2}^{3}\phantom{\rule{0ex}{0ex}}=\left(x-1+2\right)\left[{\left(x-1\right)}^{2}+4-2\left(x-1\right)\right]\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left[{x}^{2}+1-2x+4-2x+2\right]\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left({x}^{2}-4x+7\right)$

#### Page No 129:

(x +1)3 + (x – 1)3
$=\left(x+1+x-1\right)\left[{\left(x+1\right)}^{2}+{\left(x-1\right)}^{2}-\left(x-1\right)\left(x+1\right)\right]\phantom{\rule{0ex}{0ex}}=\left(2x\right)\left[{\left(x+1\right)}^{2}+{\left(x-1\right)}^{2}-\left({x}^{2}-1\right)\right]\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+1+2x+{x}^{2}+1-2x-{x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+3\right)$

#### Page No 129:

(2a +1)3 + (a – 1)3
$=\left(2a+1+a-1\right)\left[{\left(2a+1\right)}^{2}+{\left(a-1\right)}^{2}-\left(2a+1\right)\left(a-1\right)\right]\phantom{\rule{0ex}{0ex}}=\left(3a\right)\left[4{a}^{2}+1+4a+{a}^{2}+1-2a-2{a}^{2}+2a-a+1\right]\phantom{\rule{0ex}{0ex}}=3a\left[3{a}^{2}+3a+3\right]=9a\left({a}^{2}+a+1\right)$

#### Page No 129:

8(x +y)3 – 27(x – y)3
$={\left[2\left(x+y\right)\right]}^{3}-{\left[3\left(x-y\right)\right]}^{3}\phantom{\rule{0ex}{0ex}}=\left(2x+2y-3x+3y\right)\left[4{\left(x+y\right)}^{2}+9{\left(x-y\right)}^{2}+6\left({x}^{2}-{y}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(-x+5y\right)\left[4\left({x}^{2}+{y}^{2}+2xy\right)+9\left({x}^{2}+{y}^{2}-2xy\right)+6\left({x}^{2}-{y}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(-x+5y\right)\left[4{x}^{2}+4{y}^{2}+8xy+9{x}^{2}+9{y}^{2}-18xy+6{x}^{2}-6{y}^{2}\right]\phantom{\rule{0ex}{0ex}}=\left(-x+5y\right)\left(19{x}^{2}+7{y}^{2}-10xy\right)$

#### Page No 129:

(x +2)3 + (x – 2)3
$=\left(x+2+x-2\right)\left[{\left(x+2\right)}^{2}+{\left(x-2\right)}^{2}-\left({x}^{2}-4\right)\right]\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+4+4x+{x}^{2}+4-4x-{x}^{2}+4\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+12\right)\phantom{\rule{0ex}{0ex}}$

#### Page No 129:

(x + 2)3 – (x – 2)3
$=\left(x+2-x+2\right)\left[{\left(x+2\right)}^{2}+{\left(x-2\right)}^{2}+\left({x}^{2}-4\right)\right]\phantom{\rule{0ex}{0ex}}=4\left[{x}^{2}+4+4x+{x}^{2}+4-4x+{x}^{2}-4\right]\phantom{\rule{0ex}{0ex}}=4\left(3{x}^{2}+4\right)$

#### Page No 129:

$\frac{{\left(0.85\right)}^{3}+{\left(0.15\right)}^{3}}{{\left(0.85\right)}^{2}-0.85×0.15+{\left(0.15\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(0.85+0.15\right)\left({\left(0.85\right)}^{2}-0.85×0.15+{\left(0.15\right)}^{2}\right)}{{\left(0.85\right)}^{2}-0.85×0.15+{\left(0.15\right)}^{2}}\phantom{\rule{0ex}{0ex}}=0.85+0.15=1:\mathrm{RHS}$
Thus, LHS = RHS

#### Page No 129:

$\frac{59×59×59-9×9×9}{59×59+59×9+9×9}\phantom{\rule{0ex}{0ex}}=\frac{{\left(59\right)}^{3}-{9}^{3}}{{59}^{2}+59×9+{9}^{2}}\phantom{\rule{0ex}{0ex}}$

Thus, LHS=RHS

#### Page No 136:

(x – y − z) (x2 + y2 + z2 + xy – yz + xz)

#### Page No 136:

$\left(3x-5y+4\right)\left(9{x}^{2}+25{y}^{2}+15xy-20y+12x+16\right)\phantom{\rule{0ex}{0ex}}=\left(3x+\left(-5y\right)+4\right)\left(9{x}^{2}+25{y}^{2}+16+15xy-20y+12x\right)$

$\left(3x+\left(-5y\right)+4\right)\left(9{x}^{2}+25{y}^{2}+16+15xy-20y+12x\right)\phantom{\rule{0ex}{0ex}}={\left(3x\right)}^{3}+{\left(-5y\right)}^{3}+{4}^{3}-3×3x\left(-5y\right)\left(4\right)\phantom{\rule{0ex}{0ex}}=27{x}^{3}-125{y}^{3}+64+180xy$

#### Page No 136:

#### Page No 137:

${\left(a-b\right)}^{3}+{\left(b-c\right)}^{3}+{\left(c-a\right)}^{3}$

#### Page No 137:

${\left(a-3b\right)}^{3}+{\left(3b-c\right)}^{3}+{\left(c-a\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(a-3b+3b-c+c-a\right)\left[{\left(a-3b\right)}^{2}+{\left(3b-c\right)}^{2}+{\left(c-a\right)}^{2}-\left(a-3b\right)\left(3b-c\right)-\left(3b-c\right)\left(c-a\right)-\left(c-a\right)\left(a-3b\right)\right]\phantom{\rule{0ex}{0ex}}+3\left(a-3b\right)\left(3b-c\right)\left(c-a\right)\phantom{\rule{0ex}{0ex}}=0+3\left(a-3b\right)\left(3b-c\right)\left(c-a\right)\phantom{\rule{0ex}{0ex}}=3\left(a-3b\right)\left(3b-c\right)\left(c-a\right)$

#### Page No 137:

(i) (–12)+ 7+ 53

${\left(-12\right)}^{3}+{7}^{3}+{5}^{3}\phantom{\rule{0ex}{0ex}}=\left(-12+7+5\right)\left[{\left(-12\right)}^{2}+{7}^{2}+{5}^{2}-7\left(-12\right)-35+60\right]+3\left(-12\right)×35\phantom{\rule{0ex}{0ex}}=0-1260=-1260$

(ii) (28)3 + (–15)3 + (–13)3

${\left(28\right)}^{3}+{\left(-15\right)}^{3}+{\left(-13\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(28-15-13\right)\left[{\left(28\right)}^{2}+{\left(-15\right)}^{2}+{\left(-13\right)}^{2}-28\left(-15\right)-\left(-15\right)\left(-13\right)-28\left(-13\right)\right]+3×28\left(-15\right)\left(-13\right)\phantom{\rule{0ex}{0ex}}=0+16380=16380$

#### Page No 137:

${\left(a+b+c\right)}^{3}={\left[\left(a+b\right)+c\right]}^{3}={\left(a+b\right)}^{3}+{c}^{3}+3\left(a+b\right)c\left(a+b+c\right)\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}={a}^{3}+{b}^{3}+3ab\left(a+b\right)+{c}^{3}+3\left(a+b\right)c\left(a+b+c\right)\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3\left(a+b\right)\left[ab+ca+cb+{c}^{2}\right]\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 137:

$a+b+c=0⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

Thus, we have:
$\left(\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}\right)=\frac{{a}^{3}+{b}^{3}+{c}^{3}}{abc}$
$=\frac{3abc}{abc}\phantom{\rule{0ex}{0ex}}=3$

#### Page No 137:

a + b + c = 9
$⇒{\left(a+b+c\right)}^{2}={9}^{2}=81\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}+{c}^{2}+2\left(ab+bc+ca\right)=81\phantom{\rule{0ex}{0ex}}⇒35+2\left(ab+bc+ca\right)=81\phantom{\rule{0ex}{0ex}}⇒\left(ab+bc+ca\right)=23$
We know,
(a+ b3 + c3 – 3abc) = $\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)$
$=\left(9\right)\left(35-23\right)\phantom{\rule{0ex}{0ex}}=108$

(c) 2

#### Page No 138:

(249)2 – (248)2
We know

Hence, the correct answer is option (d).

#### Page No 138:

(c) 0

$⇒x$2 + y2 = $-$xy
x2 + y2 + xy = 0

Thus, we have:
$\left({x}^{3}-{y}^{3}\right)=\left(x-y\right)\left({x}^{2}+{y}^{2}+xy\right)$
$=\left(x-y\right)×0\phantom{\rule{0ex}{0ex}}=0$

#### Page No 138:

(d) 3abc

$⇒{\left(a+b\right)}^{3}={\left(-c\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+3ab\left(a+b\right)=-{c}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+3ab\left(-c\right)=-{c}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

#### Page No 139:

Hence, the correct answer is option (c).

#### Page No 139:

(x + 3)3
$={x}^{3}+{3}^{3}+9x\left(x+3\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+27+9{x}^{2}+27x$
So, the coefficient of x in (x + 3)is 27.
Hence, the correct answer is option (d).

#### Page No 139:

(x + y)3 – (xy3)
$={x}^{3}+{y}^{3}+3xy\left(x+y\right)-\left({x}^{3}+{y}^{3}\right)\phantom{\rule{0ex}{0ex}}=3xy\left(x+y\right)$
Thus, the factors of (x + y)3 – (xy3) are 3xy and (x + y).
Hence, the correct answer is option (d).

#### Page No 139:

$\left(25{x}^{2}-1\right)+{\left(1+5x\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(5x-1\right)\left(5x+1\right)+{\left(1+5x\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(5x+1\right)\left[\left(5x-1\right)+\left(1+5x\right)\right]\phantom{\rule{0ex}{0ex}}=\left(5x+1\right)\left(10x\right)$
So, the factors of $\left(25{x}^{2}-1\right)+{\left(1+5x\right)}^{2}$ are (5x + 1) and 10x
Hence, the correct answer is option (d).

(b) 5

#### Page No 139:

(b) m = 7, n = −18

Let:
$p\left(x\right)={x}^{3}+10{x}^{2}+mx+n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Now,
$x+2=0⇒x=-2$
(x + 2) is a factor of p(x).
So, we have p($-$2)=0

Now,
$x-1=0⇒x=1$
Also,
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0

By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

#### Page No 139:

(b) 9984

$104×96=\left(100+4\right)\left(100-4\right)$
$={100}^{2}-{4}^{2}\phantom{\rule{0ex}{0ex}}=\left(10000-16\right)\phantom{\rule{0ex}{0ex}}=9984$

(c) 93940

(b) 39951

#### Page No 139:

(a) (2a + b + 2)2

#### Page No 139:

(c) (x − 7)(x + 3)

${x}^{2}-4x-21={x}^{2}-7x+3x-21$
$=x\left(x-7\right)+3\left(x-7\right)\phantom{\rule{0ex}{0ex}}=\left(x-7\right)\left(x+3\right)$

#### Page No 139:

(c) (2x + 3) (2x − 1)

$4{x}^{2}+4x-3=4{x}^{2}+6x-2x-3$
$=2x\left(2x+3\right)-1\left(2x+3\right)\phantom{\rule{0ex}{0ex}}=\left(2x+3\right)\left(2x-1\right)$

#### Page No 139:

(b) (2x + 5)(3x + 1)

$6{x}^{2}+17x+5=6{x}^{2}+15x+2x+5$
$=3x\left(2x+5\right)+1\left(2x+5\right)\phantom{\rule{0ex}{0ex}}=\left(2x+5\right)\left(3x+1\right)$

#### Page No 139:

(c) x3 − 2x2 − x − 2

Let:
$f\left(x\right)={x}^{3}-2{x}^{2}+x+2$
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:

Hence, (x + 1) is not a factor of $f\left(x\right)={x}^{3}-2{x}^{2}+x+2$.

Now,
Let:
$f\left(x\right)={x}^{3}+2{x}^{2}+x-2$

By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:

Hence, (x + 1) is not a factor of $f\left(x\right)={x}^{3}+2{x}^{2}+x-2$.

Now,
Let:
$f\left(x\right)={x}^{3}+2{x}^{2}-x-2$

By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:

Hence, (x + 1) is a factor of $f\left(x\right)={x}^{3}+2{x}^{2}-x-2$.

#### Page No 140:

(d) (3x + 2)(x2 + 1)

$3{x}^{3}+2{x}^{2}+3x+2={x}^{2}\left(3x+2\right)+1\left(3x+2\right)$
$=\left(3x+2\right)\left({x}^{2}+1\right)$

#### Page No 140:

(d) 3

$a+b+c=0⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

Thus, we have:
$\left(\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}\right)=\frac{{a}^{3}+{b}^{3}+{c}^{3}}{abc}$
$=\frac{3abc}{abc}\phantom{\rule{0ex}{0ex}}=3$

#### Page No 140:

(a) 108

${x}^{3}+{y}^{3}+{z}^{3}-3xyz=\left(x+y+z\right)\left({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx\right)\phantom{\rule{0ex}{0ex}}$
$=\left(x+y+z\right)\left[{\left(x+y+z\right)}^{2}-3\left(xy+yz+zx\right)\right]\phantom{\rule{0ex}{0ex}}=9×\left(81-3×23\right)\phantom{\rule{0ex}{0ex}}=9×12\phantom{\rule{0ex}{0ex}}=108$

#### Page No 140:

$⇒a$2 + b2 = $-$ab
$⇒a$2 + b2 + ab = 0
$\left({a}^{3}-{b}^{3}\right)=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$
$=\left(a-b\right)×0\phantom{\rule{0ex}{0ex}}=0$