Rs Aggarwal 2020 2021 Solutions for Class 8 Maths 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 Maths Cubes And Cube Roots Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Page No 64:

(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.

#### Page No 64:

(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.

#### Page No 65:

(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

#### Page No 65:

(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.

#### Page No 65:

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 =

#### Page No 65:

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 = $×$ ${7}^{3}$=

#### Page No 65:

1323

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

#### Page No 65:

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.

#### Page No 65:

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:
$5×5=25$

#### Page No 65:

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.

#### Page No 66:

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

#### Page No 66:

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

#### Page No 66:

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

#### Page No 66:

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

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

#### Page No 67:

$\sqrt[3]{64}$
By prime factorisation:
64 = $2×2×2×2×2×2$
= $\left(2×2×2\right)×\left(2×2×2\right)$

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

#### Page No 67:

$\sqrt[3]{343}$
By prime factorisation:
343 = $7×7×7$
= ( $7×7×7$ )

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

#### Page No 67:

$\sqrt[3]{729}$
By prime factorisation:

729 = $3×3×3×3×3×3$
= $\left(3×3×3\right)×\left(3×3×3\right)$

$\sqrt[3]{729}$ =

#### Page No 67:

$\sqrt[3]{1728}$
By prime factorisation:

1728 = $2×2×2×2×2×2×3×3×3$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(3×3×3\right)={2}^{3}×{2}^{3}×{3}^{3}$

$\sqrt[3]{1728}$ =

#### Page No 67:

$\sqrt[3]{9261}$
By prime factorisation:

9261 = $3×3×3×7×7×7$
= $\left(3×3×3\right)×\left(7×7×7\right)={3}^{3}×{7}^{3}$

$\sqrt[3]{9261}$ =

#### Page No 67:

$\sqrt[3]{4096}$
By prime factorisation:

4096 = $2×2×2×2×2×2×2×2×2×2×2×2$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(2×2×2\right)×\left(2×2×2\right)\phantom{\rule{0ex}{0ex}}={2}^{3}×{2}^{3}×{2}^{3}×{2}^{3}$

$\sqrt[3]{4096}$ =

#### Page No 67:

$\sqrt[3]{8000}$
By prime factorisation:

8000 = $2×2×2×2×2×2×5×5×5$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(5×5×5\right)$

$\sqrt[3]{8000}$ =

#### Page No 67:

$\sqrt[3]{3375}$
By prime factorisation:

3375 = $3×3×3×5×5×5$
= $\left(3×3×3\right)×\left(5×5×5\right)$

$\sqrt[3]{3375}$ =

#### Page No 68:

$\sqrt[3]{-216}$
By prime factorisation:

216 = $2×2×2×3×3×3$
= $\left(2×2×2\right)×\left(3×3×3\right)$
$\sqrt[3]{-216}$ =

$\sqrt[3]{-216}$ =  $-\left(\sqrt[3]{216}\right)=-6$

#### Page No 68:

$\sqrt[3]{-512}$
By prime factorisation:

$\sqrt[3]{512}$ = $2×2×2×2×2×2×2×2×2$
= $\left(2×2×2\right)×\left(2×2×2\right)×\left(2×2×2\right)$
$\sqrt[3]{-512}$ =

$\sqrt[3]{-512}$$-\left(\sqrt[3]{512}\right)=-8$

#### Page No 68:

$\sqrt[3]{-1331}$
By prime factorisation:
$\sqrt[3]{1331}$ = $\sqrt[3]{11×11×11}$

$\sqrt[3]{-1331}$=
$\sqrt[3]{-1331}=-\left(\sqrt[3]{1331}\right)=-11$

#### Page No 68:

$\sqrt[3]{\frac{27}{64}}$
By prime factorisation:

$\sqrt[3]{\frac{27}{64}}$ = $\frac{\sqrt[3]{27}}{\sqrt[3]{64}}$= $\frac{\sqrt[3]{\left(3×3×3\right)}}{\sqrt[3]{\left(2×2×2\right)×\left(2×2×2\right)}}$= $\frac{\sqrt[3]{\left(3×3×3\right)}}{\sqrt[3]{\left(4×4×4\right)}}=\frac{3}{4}$
$\sqrt[3]{\frac{27}{64}}$= $\frac{3}{4}$

#### Page No 68:

$\sqrt[3]{\frac{125}{216}}$
By prime factorisation:

$\sqrt[3]{\frac{125}{216}}$  = $\frac{\sqrt[3]{5×5×5}}{\sqrt[3]{\left(2×2×2\right)×\left(3×3×3\right)}}$ = $\frac{\sqrt[3]{5×5×5}}{\sqrt[3]{\left(6×6×6\right)}}=\frac{5}{6}$

$\sqrt[3]{\frac{125}{216}}$ = $\frac{5}{6}$

#### Page No 68:

$\sqrt[3]{\frac{-27}{125}}$

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

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

#### Page No 68:

$\sqrt[3]{\frac{-64}{343}}$
On factorisation:

$\sqrt[3]{\frac{64}{343}}$= $\sqrt[3]{\frac{2×2×2×2×2×2}{7×7×7}}$
$\sqrt[3]{\frac{-64}{343}}$= $\frac{-4}{7}$

#### Page No 68:

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

#### Page No 68:

$\sqrt[3]{\frac{729}{1000}}$

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

#### Page No 68:

$\sqrt[3]{\frac{-512}{343}}$
By factorisation:

$\sqrt[3]{\frac{512}{343}}$= $\frac{\sqrt[3]{8×8×8}}{\sqrt[3]{7×7×7}}$
$\sqrt[3]{\frac{-512}{343}}$= $\frac{-8}{7}$

#### Page No 68:

(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.

#### Page No 68:

(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.

#### Page No 68:

(c) 8

$\sqrt[3]{512}$ =
$\sqrt[3]{512}$ =

Hence, the cube root of 512 is 8.

#### Page No 68:

(c) 20

Hence, the cube root of $\sqrt[3]{125×64}$ is 20.

#### Page No 68:

(b) $\frac{4}{7}$
$\sqrt[3]{\frac{64}{343}}$ =
$\sqrt[3]{\frac{64}{343}}$ = $\frac{4}{7}$
$\sqrt[3]{\frac{64}{343}}$ = $\frac{4}{7}$

#### Page No 68:

(b) $\frac{-8}{9}$
$\sqrt[3]{\frac{-512}{729}}$=
$\sqrt[3]{\frac{-512}{729}}$ = $\frac{-8}{9}$
$\sqrt[3]{\frac{-512}{729}}$ = $\frac{-8}{9}$

#### Page No 68:

(c) 9

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

#### Page No 68:

(a) 3

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

#### Page No 68:

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

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

#### Page No 68:

(c) 0.512

= $0.512$

#### Page No 70:

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

#### Page No 70:

$\sqrt[3]{4096}$

By prime factorisation method:

$\sqrt[3]{4096}$ =
$\sqrt[3]{4096}$ = .

$\sqrt[3]{4096}$ = $16$

#### Page No 70:

$\sqrt[3]{216×343}$
By prime factorisation:

$\sqrt[3]{216×343}$ =
$\sqrt[3]{216×343}$ =

$\sqrt[3]{216×343}$ = $42$

#### Page No 70:

$\sqrt[3]{\frac{-64}{125}}$
By prime factorisation method:

$\sqrt[3]{\frac{-64}{125}}$ =
$\sqrt[3]{\frac{-64}{125}}$ = $\frac{-4}{5}$
$\sqrt[3]{\frac{-64}{125}}$ = $\frac{-4}{5}$

#### Page No 70:

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

#### Page No 70:

(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.

(c) 24

=
=
=
=  24

#### Page No 70:

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

$\sqrt[3]{\frac{-343}{729}}$ = $\frac{-7}{9}$

#### Page No 70:

(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.

#### Page No 70:

(b) $\frac{4}{5}$

Resolving the numerator and the denominator into prime factors:

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

#### Page No 70:

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

#### Page No 70:

(i) $\sqrt[3]{b}$

(ii) $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$

(iii) $-\sqrt[3]{x}$

(iv) 0.125

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