Rs Aggarwal 2017 Solutions for Class 8 Math Chapter 4 Cubes And Cube Roots are provided here with simple step-by-step explanations. These solutions for Cubes And Cube Roots are extremely popular among Class 8 students for Math Cubes And Cube Roots Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 8 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Question 1:

Evaluate:
(i) (8)3
(ii) (15)3
(iii) (21)3
(iv) (60)3

(i) (8)3 = $\left(8×8×8\right)$= 512.
Thus, the cube of 8 is 512.
(ii) (15)3 = $\left(15×15×15\right)$= 3375.
Thus, the cube of 15 is 3375.
(iii) (21)3 = $\left(21×21×21\right)$= 9261.
Thus, the cube of 21 is 9261.
(iv) (60)3 = $\left(60×60×60\right)$= 216000.
Thus, the cube of 60 is 216000.

#### Question 2:

Evaluate:
(i) (1.2)3
(ii) (3.5)3
(iii) (0.8)3
(iv) (0.05)3

(i) (1.2)3 = $\left(1.2×1.2×1.2\right)$= 1.728
Thus, the cube of 1.2 is 1.728.
(ii) (3.5)3= $\left(3.5×3.5×3.5\right)$= 42.875
Thus, the cube of 3.5 is 42.875.
(iii) (0.8)3= $\left(0.8×0.8×0.8\right)$= 0.512
Thus, the cube of 0.8 is 0.512.
(iv) (0.05)3= $\left(0.05×0.05×0.05\right)$= 0.000125
Thus, the cube of 0.05 is 0.000125.

#### Question 3:

Evaluate:
(i) ${\left(\frac{4}{7}\right)}^{3}$
(ii) ${\left(\frac{10}{11}\right)}^{3}$
(iii) ${\left(\frac{1}{15}\right)}^{3}$
(iv) ${\left(1\frac{3}{10}\right)}^{3}$

(i) ${\left(\frac{4}{7}\right)}^{3}$= $\left(\frac{4}{7}×\frac{4}{7}×\frac{4}{7}\right)$= $\left(\frac{64}{343}\right)$
Thus, the cube of $\left(\frac{4}{7}\right)$ is $\left(\frac{64}{343}\right)$.

(ii)
${\left(\frac{10}{11}\right)}^{3}$= $\left(\frac{10}{11}×\frac{10}{11}×\frac{10}{11}\right)$= $\left(\frac{1000}{1331}\right)$
Thus, the cube of $\left(\frac{10}{11}\right)$ is $\left(\frac{1000}{1331}\right)$.(47)3
(iii)
${\left(\frac{1}{15}\right)}^{3}$= $\left(\frac{1}{15}×\frac{1}{15}×\frac{1}{15}\right)$= $\left(\frac{1}{3375}\right)$
Thus, the cube of $\left(\frac{1}{15}\right)$ is $\left(\frac{1}{3375}\right)$
(iv)
${\left(1\frac{3}{10}\right)}^{3}$${\left(\frac{13}{10}\right)}^{3}$= $\left(\frac{13}{10}×\frac{13}{10}×\frac{13}{10}\right)$= $\left(\frac{2197}{1000}\right)$
Thus, the cube of $\left(1\frac{3}{10}\right)$ is $\left(\frac{2197}{1000}\right)$. (47)3

#### Question 4:

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
(i) 125
(ii) 243
(iii) 343
(iv) 256
(v) 8000
(vi) 9261
(vii) 5324
(viii) 3375

(i) 125
Resolving 125 into prime factors:
125 = 5$×$5$×$5
Here, one triplet is formed, which is ${5}^{3}$. Hence, 125 can be expressed as the product of the triplets of 5.
Therefore, 125 is a perfect cube.

(ii) 243 is not a perfect cube.

(iii) 343
Resolving 125 into prime factors:
343 = 7$×$7$×$7
Here, one triplet is formed, which is ${7}^{3}$. Hence, 343 can be expressed as the product of the triplets of 7.
Therefore, 343 is a perfect cube.

(iv) 256 is not a perfect cube.

(v) 8000
Resolving 8000 into prime factors:
8000 = 2$×$2$×$2$×$2$×$2$×$2$×$5$×$5$×$5
Here, three triplets are formed, which are 23, 23 and 53. Hence, 8000 can be expressed as the product of the triplets of 2, 2 and 5, i.e. $×$$×$ = .
Therefore, 8000 is a perfect cube.

(vi) 9261
Resolving 9261 into prime factors:
9261 = 3$×$3$×$3$×$7$×$7$×$7
Here, two triplets are formed, which are and ${7}^{3}$. Hence, 9261 can be expressed as the product of the triplets of 3 and 7, i.e. $×$ ${7}^{3}$= .
Therefore, 9261 is a perfect cube.

(vii) 5324 is not a perfect cube.

(viii) 3375 .
Resolving 3375 into prime factors:
3375 = 3$×$3$×$3$×$5$×$5$×$5.
Here, two triplets are formed, which are and ${5}^{3}$. Hence, 3375 can be expressed as the product of the triplets of 3 and 5, i.e. $×$ ${5}^{3}$= .
Therefore, 3375 is a perfect cube.

#### Question 5:

Which of the following are the cubes of even numbers?
(i) 216
(ii) 729
(iii) 512
(iv) 3375
(v) 1000

The cubes of even numbers are always even. Therefore, 216, 512 and 1000 are the cubes of even numbers.

216 = 2$×$2$×$2$×$3$×$3$×$3 =
512 =
1000 =

#### Question 6:

Which of the following are the cubes of odd numbers?
(i) 125
(ii) 343
(iii) 1728
(iv) 4096
(v) 9261

The cube of an odd number is an odd number. Therefore, 125, 343 and 9261 are the cubes of odd numbers.

125 = 5$×$5$×$5 = ${5}^{3}$

343 = 7$×$7$×$7 = ${7}^{3}$

9261 = 3$×$3$×$3$×$7$×$7$×$7 = $×$ ${7}^{3}$=

#### Question 7:

Find the smallest number by which 1323 must be multiplied so that the product is a perfect cube.

1323 1323 = $3×3×3×7×7$.
To make it a perfect cube, it has to be multiplied by 7.

#### Question 8:

Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.

2560

2560 can be expressed as the product of prime factors in the following manner: 2560 = $2×2×2×2×2×2×2×2×2×5$

To make this a perfect square, we have to multiply it by 5$×$5.
Therefore, 2560 should be multiplied by 25 so that the product is a perfect cube.

#### Question 9:

What is the smallest number by which 1600 must be divided so that the quotient is a perfect cube?

1600

1600 can be expressed as the product of prime factors in the following manner: 1600 = $2×2×2×2×2×2×5×5$

Therefore, to make the quotient a perfect cube, we have to divide 1600 by:
$5×5=25$

#### Question 10:

Find the smallest number by which 8788 must be divided so that the quotient is a perfect cube. 8788
8788 can be expressed as the product of prime factors as $2×2×13×13×13$.
Therefore, 8788 should be divided by 4, i.e. ($2×2$), so that the quotient is a perfect cube.

#### Question 1:

Find the value of using the short-cut method:
(25)3

${\left(25\right)}^{3}$
Here, a = 2 and b = 5

Using the formula ${a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$:

 4 $×$ 2 4 $×$  15 25 $×$  6 25 $×$5 8  $+$7 60   $+$ 16 150 $+$  12 125 15 76 162

${\left(25\right)}^{3}$ = 15625

#### Question 2:

Find the value of using the short-cut method:
(47)3

${\left(47\right)}^{3}$
Here, a = 4 and b = 7

Using the formula ${a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$:

 16 $×$ 4 16 $×$  21 49 $×$  12 49 $×$7 64  $+$39 336   $+$   62 588 $+$  34 343 103 398 622

${\left(47\right)}^{3}$ = 103823

#### Question 3:

Find the value of using the short-cut method:
(68)3

${\left(68\right)}^{3}$
Here, a = 6 and b = 8

Using the formula ${a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$:

 36 $×$ 6 36 $×$  24 64 $×$  18 64 $×$8 216  $+$  98 864   $+$ 120 1152 $+$    51 512 314 984 1203

${\left(68\right)}^{3}$ = 314432

#### Question 4:

Find the value of using the short-cut method:
(84)3

${\left(84\right)}^{3}$

Here, a = 8 and b = 4

Using the formula ${a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$:

 64 $×$   8 64 $×$  12 16 $×$  24 16 $×$ 4 512  $+$  80 768   $+$ 39 384 $+$    6 64 592 807 390

${\left(84\right)}^{3}$ = 592704

#### Question 1:

Evaluate:
$\sqrt{64}$

$\sqrt{64}$
By prime factorisation:
64 = $2×2×2×2×2×2$
= $\left(2×2×2\right)×\left(2×2×2\right)$

$\sqrt{64}=\sqrt{\left(2{\right)}^{3}×\left(2{\right)}^{3}}=\left(2×2\right)=4$

#### Question 2:

Evaluate:
$\sqrt{343}$

$\sqrt{343}$
By prime factorisation:
343 = $7×7×7$
= ( $7×7×7$ )

$\sqrt{343}=\sqrt{{7}^{3}}=7$

#### Question 3:

Evaluate:
$\sqrt{729}$

$\sqrt{729}$
By prime factorisation: 729 = $3×3×3×3×3×3$
= $\left(3×3×3\right)×\left(3×3×3\right)$

$\sqrt{729}$ =

#### Question 4:

Evaluate:
$\sqrt{1728}$

$\sqrt{1728}$
By prime factorisation: 1728 = $2×2×2×2×2×2×3×3×3$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(3×3×3\right)={2}^{3}×{2}^{3}×{3}^{3}$

$\sqrt{1728}$ =

#### Question 5:

Evaluate:
$\sqrt{9261}$

$\sqrt{9261}$
By prime factorisation: 9261 = $3×3×3×7×7×7$
= $\left(3×3×3\right)×\left(7×7×7\right)={3}^{3}×{7}^{3}$

$\sqrt{9261}$ =

#### Question 6:

Evaluate:
$\sqrt{4096}$

$\sqrt{4096}$
By prime factorisation: 4096 = $2×2×2×2×2×2×2×2×2×2×2×2$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(2×2×2\right)×\left(2×2×2\right)\phantom{\rule{0ex}{0ex}}={2}^{3}×{2}^{3}×{2}^{3}×{2}^{3}$

$\sqrt{4096}$ =

#### Question 7:

Evaluate:
$\sqrt{8000}$

$\sqrt{8000}$
By prime factorisation: 8000 = $2×2×2×2×2×2×5×5×5$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(5×5×5\right)$

$\sqrt{8000}$ =

#### Question 8:

Evaluate:
$\sqrt{3375}$

$\sqrt{3375}$
By prime factorisation: 3375 = $3×3×3×5×5×5$
= $\left(3×3×3\right)×\left(5×5×5\right)$

$\sqrt{3375}$ =

#### Question 9:

Evaluate:
$\sqrt{-216}$

$\sqrt{-216}$
By prime factorisation: 216 = $2×2×2×3×3×3$
= $\left(2×2×2\right)×\left(3×3×3\right)$
$\sqrt{-216}$ =

$\sqrt{-216}$ =  $-\left(\sqrt{216}\right)=-6$

#### Question 10:

Evaluate:
$\sqrt{-512}$

$\sqrt{-512}$
By prime factorisation: $\sqrt{512}$ = $2×2×2×2×2×2×2×2×2$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(2×2×2\right)$
$\sqrt{-512}$ =

$\sqrt{-512}$$-\left(\sqrt{512}\right)=-8$

#### Question 11:

Evaluate:
$\sqrt{-1331}$

$\sqrt{-1331}$
By prime factorisation:
$\sqrt{1331}$ = $\sqrt{11×11×11}$ $\sqrt{-1331}$=
$\sqrt{-1331}=-\left(\sqrt{1331}\right)=-11$

#### Question 12:

Evaluate:
$\sqrt{\frac{27}{64}}$

$\sqrt{\frac{27}{64}}$
By prime factorisation:  $\sqrt{\frac{27}{64}}$ = $\frac{\sqrt{27}}{\sqrt{64}}$= $\frac{\sqrt{\left(3×3×3\right)}}{\sqrt{\left(2×2×2\right)×\left(2×2×2\right)}}$= $\frac{\sqrt{\left(3×3×3\right)}}{\sqrt{\left(4×4×4\right)}}=\frac{3}{4}$
$\sqrt{\frac{27}{64}}$= $\frac{3}{4}$

#### Question 13:

Evaluate:
$\sqrt{\frac{125}{216}}$

$\sqrt{\frac{125}{216}}$
By prime factorisation:  $\sqrt{\frac{125}{216}}$  = $\frac{\sqrt{5×5×5}}{\sqrt{\left(2×2×2\right)×\left(3×3×3\right)}}$ = $\frac{\sqrt{5×5×5}}{\sqrt{\left(6×6×6\right)}}=\frac{5}{6}$

$\sqrt{\frac{125}{216}}$ = $\frac{5}{6}$

#### Question 14:

Evaluate:
$\sqrt{\frac{-27}{125}}$

$\sqrt{\frac{-27}{125}}$  By factorisation:
$\sqrt{\frac{27}{125}}$= $\sqrt{\frac{3×3×3}{5×5×5}}$

$\sqrt{\frac{-27}{125}}$= $\frac{-3}{5}$

#### Question 15:

Evaluate:
$\sqrt{\frac{-64}{343}}$

$\sqrt{\frac{-64}{343}}$
On factorisation:  $\sqrt{\frac{64}{343}}$= $\sqrt{\frac{2×2×2×2×2×2}{7×7×7}}$
$\sqrt{\frac{-64}{343}}$= $\frac{-4}{7}$

#### Question 16:

Evaluate:
$\sqrt{64×729}$

$\sqrt{64×729}$
$\sqrt{64×729}$ = $\sqrt{64}×\sqrt{729}$
= $\sqrt{4×4×4}$ $×\sqrt{\left(3×3×3\right)×\left(3×3×3\right)}$
= $\sqrt{4×4×4}$ $×\sqrt{\left(9×9×9\right)}$
$\sqrt{64×729}$ = $\left(4\right)×\left(9\right)$= 36

#### Question 17:

Evaluate:
$\sqrt{\frac{729}{1000}}$

$\sqrt{\frac{729}{1000}}$  On factorisation:
$\sqrt{\frac{729}{1000}}$$\frac{\sqrt{\left(3×3×3\right)×\left(3×3×3\right)}}{\sqrt{\left(2×2×2\right)×\left(5×5×5\right)}}$= $\frac{\sqrt{9×9×9}}{\sqrt{10×10×10}}$
$\sqrt{\frac{729}{1000}}$= $\frac{9}{10}$

#### Question 18:

Evaluate:
$\sqrt{\frac{-512}{343}}$

$\sqrt{\frac{-512}{343}}$
By factorisation:  $\sqrt{\frac{512}{343}}$= $\frac{\sqrt{8×8×8}}{\sqrt{7×7×7}}$
$\sqrt{\frac{-512}{343}}$= $\frac{-8}{7}$

#### Question 1:

Which of the following numbers is a perfect cube?
(a) 141
(b) 294
(c) 216
(d) 496

(a)
141 is not a perfect cube.

(b)
294 is not a perfect cube.

(c) (✓)
216 is a perfect cube.
216 =

(d)
496 is not a perfect cube.

#### Question 2:

Which of the following numbers is a perfect cube?
(a) 1152
(b) 1331
(c) 2016
(d) 739

(a)
1152 = .
Hence, 1152 is not a perfect cube.

(b) (✓)
1331 =
Hence, 1331 is a perfect cube.

(c)
2016 =
Hence, 2016 is not a perfect cube.

(d)
739 is not a perfect cube.

#### Question 3:

$\sqrt{512}=?$
(a) 6
(b) 7
(c) 8
(d) 9

(c) 8

$\sqrt{512}$ =
$\sqrt{512}$ =

Hence, the cube root of 512 is 8.

#### Question 4:

$\sqrt{125×64}=?$
(a) 100
(b) 40
(c) 20
(d) 30

(c) 20

Hence, the cube root of $\sqrt{125×64}$ is 20.

#### Question 5:

$\sqrt{\frac{64}{343}}=?$
(a) $\frac{4}{9}$
(b) $\frac{4}{7}$
(c) $\frac{8}{7}$
(d) $\frac{8}{21}$

(b) $\frac{4}{7}$
$\sqrt{\frac{64}{343}}$ =
$\sqrt{\frac{64}{343}}$ = $\frac{4}{7}$
$\sqrt{\frac{64}{343}}$ = $\frac{4}{7}$

#### Question 6:

$\sqrt{\frac{-512}{729}}=?$
(a) $\frac{-7}{9}$
(b) $\frac{-8}{9}$
(c) $\frac{7}{9}$
(d) $\frac{8}{9}$

(b) $\frac{-8}{9}$
$\sqrt{\frac{-512}{729}}$=
$\sqrt{\frac{-512}{729}}$ = $\frac{-8}{9}$
$\sqrt{\frac{-512}{729}}$ = $\frac{-8}{9}$

#### Question 7:

By what least number should 648 be multiplied to get a perfect cube?
(a) 3
(b) 6
(c) 9
(d) 8

(c) 9 648 =
Therefore, to get a perfect cube, we need to multiply 648 by 9, i.e. $\left(3×3\right)$.

#### Question 8:

By what least number should 1536 be divided to get a perfect cube?
(a) 3
(b) 4
(c) 6
(d) 8

(a) 3 1536 =
Therefore, to get a perfect cube, we need to divide 1536 by 3.

#### Question 9:

${\left(1\frac{3}{10}\right)}^{3}=?$
(a) $1\frac{27}{1000}$
(b) $2\frac{27}{1000}$
(c) $2\frac{197}{1000}$
(d) none of these

(c) $2\frac{197}{1000}$

${\left(1\frac{3}{10}\right)}^{3}$ = $2\frac{197}{1000}$

#### Question 10:

(0.8)3 = ?
(a) 51.2
(b) 5.12
(c) 0.512
(d) none of these

(c) 0.512

= $0.512$

#### Question 1:

Evaluate ${\left(1\frac{2}{5}\right)}^{3}.$

${\left(1\frac{2}{5}\right)}^{3}$
${\left(1\frac{2}{5}\right)}^{3}$ =
${\left(1\frac{2}{5}\right)}^{3}$ = $\frac{343}{125}$

#### Question 2:

Evaluate $\sqrt{4096}.$

$\sqrt{4096}$

By prime factorisation method:

$\sqrt{4096}$ =
$\sqrt{4096}$ = .

$\sqrt{4096}$ = $16$

#### Question 3:

Evaluate $\sqrt{216×343}.$

$\sqrt{216×343}$
By prime factorisation:

$\sqrt{216×343}$ =
$\sqrt{216×343}$ =

$\sqrt{216×343}$ = $42$

#### Question 4:

Evaluate $\sqrt{\frac{-64}{125}}.$

$\sqrt{\frac{-64}{125}}$
By prime factorisation method:

$\sqrt{\frac{-64}{125}}$ =
$\sqrt{\frac{-64}{125}}$ = $\frac{-4}{5}$
$\sqrt{\frac{-64}{125}}$ = $\frac{-4}{5}$

#### Question 5:

Mark (✓) against the correct answer
${\left(1\frac{3}{4}\right)}^{3}=?$
(a) $1\frac{27}{64}$
(b) $2\frac{27}{64}$
(c) $5\frac{23}{64}$
(d) none of these

(c) $5\frac{23}{64}$
${\left(1\frac{3}{4}\right)}^{3}$ =
${\left(1\frac{3}{4}\right)}^{3}$ = $\frac{343}{64}$ = $5\frac{23}{64}$
${\left(1\frac{3}{4}\right)}^{3}$ = $5\frac{23}{64}$

#### Question 6:

Mark (✓) against the correct answer
Which of the following numbers is a perfect cube?
(a) 121
(b) 169
(c) 196
(d) 216

(d) 216

$121=11×11\phantom{\rule{0ex}{0ex}}169=13×13\phantom{\rule{0ex}{0ex}}196=7×7×2×2$

216 =

216 = 63
Hence, 216 is a perfect cube.

#### Question 7:

Mark (✓) against the correct answer
$\sqrt{216×64}=?$
(a) 64
(b) 32
(c) 24
(d) 36

(c) 24

=
=
=
=  24

#### Question 8:

Mark (✓) against the correct answer
$\sqrt{\frac{-343}{729}}=?$
(a) $\frac{7}{9}$
(b) $\frac{-7}{9}$
(c) $\frac{-9}{7}$
(d) $\frac{9}{7}$

(b) $\frac{-7}{9}$
By prime factorisation:
$\sqrt{\frac{-343}{729}}$ =
$\sqrt{\frac{-343}{729}}$ = $\frac{\sqrt{{\left(-7\right)}^{3}}}{\sqrt{{\left(9\right)}^{3}}}$ = $\frac{-7}{9}$

$\sqrt{\frac{-343}{729}}$ = $\frac{-7}{9}$

#### Question 9:

Mark (✓) against the correct answer
By what least number should 324 be multiplied to get a perfect cube?
(a) 12
(b) 14
(c) 16
(d) 18

(d) 18 324 =

Therefore, to show that the given number is the product of three triplets, we need to multiply 324 by $\left(2×3×3\right)$.
In other words, we need to multiply 324 by 18 to make it a perfect cube.

#### Question 10:

Mark (✓) against the correct answer
$\frac{\sqrt{128}}{\sqrt{250}}=?$
(a) $\frac{3}{5}$
(b) $\frac{4}{5}$
(c) $\frac{2}{5}$
(d) none of these

(b) $\frac{4}{5}$

Resolving the numerator and the denominator into prime factors:

$\frac{\sqrt{128}}{\sqrt{250}}=\sqrt{\frac{128}{250}}=\sqrt{\frac{2×8×8}{2×5×5×5}}=\sqrt{\frac{\overline{)2}×8×8}{\overline{)2}×5×5×5}}=\sqrt{\frac{8×8}{5×5×5}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{{\left(2\right)}^{3}×{\left(2\right)}^{3}}{{\left(5\right)}^{3}}}=\frac{2×2}{5}=\frac{4}{5}$

#### Question 11:

Mark (✓) against the correct answer
Which of the following is a cube of an odd number?
(a) 216
(b) 512
(c) 343
(d) 1000

(c) 343
The cube of an odd number will always be an odd number.
Therefore, 343 is the cube of an odd number.

#### Question 12:

Fill in the blanks.
(i) $\sqrt{ab}=\left(\sqrt{a}\right)×\left(.........\right).$
(ii) $\sqrt{\frac{a}{b}}=.........$
(iii) $\sqrt{-x}=.........$
(iv) (0.5)3 = .........

(i) $\sqrt{b}$

(ii) $\frac{\sqrt{a}}{\sqrt{b}}$

(iii) $-\sqrt{x}$

(iv) 0.125

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