# ICSE Class 8 Maths Selina Solutions Chapter 4 Cubes and Cube Roots

ICSE Selina Class 8 Maths Chapter 4- Cubes and Cube-Roots explain about different steps and tricks for finding cube roots, from the very basic. Students should go through the chapter and clear their doubts so that they can practice the questions from the Class 8 Selina Maths textbook. They can cross their answers by downloading the solutions by clicking the mentioned below link. Cube of a number is obtained by multiplying the number three times, for example, the cube of 3 is obtained by 3.3.3=27. Some of the examples of perfect cubes are 1, 8, 27, 64, etc. On the other hand, the cube root is a special value that when cubed gives the original number. For eg: the cube root of 27 is 3 because when 3 is cubed you get 27.

When you start preparing for the exams make sure you download the ICSE Selina Class 8 Maths Chapter 4- Cubes and Cube-Roots Solutions.

## Download ICSE Class 8 Maths Selina Solutions PDF for Chapter 4 – Cubes and Cubes Roots

### Question 1.

Find the cube of:

(i) 7

(ii) 11

(iii) 16

(iv) 23

(v) 31

(vi) 42

(vii) 54

Solution: (i) 7

$(7)^{3}=7 \times 7 \times 7=343$

Solution: (ii) 11

$(11)^{3}=11 \times 11 \times 11=1331$

Solution: (iii) 16

$(16)^{3}=16 \times 16 \times 16=4096$

Solution: (iv) 23

$(23)^{3}=23 \times 23 \times 23=12167$

Solution: (v) 31

$(31)^{3}=31 \times 31 \times 31=29791$

Solution: (vi) 42

$(42)^{3}=42 \times 42 \times 42=74088$

Solution: (vii) 54

$(54)^{3}=54 \times 54 \times 54=157464$

### Question 2.

Find which of the following perfect cubes are:

(i) 243

(ii) 588

(iii) 1331

(iv) 24000

(v) 1728

(vi) 1938

Solution: (i) 243

Taking L.C.M.

$∵ 243=3 \times 3 \times 3 \times 3$

$=(3 \times 3 \times 3) \times 3$

$=3^{3} \times 3$

∴ 297 is not a perfect cube.

Solution: (ii) 588

Taking L.C.M.

$588=2 \times 2 \times 7 \times 7 \times 3$

∴588 is not perfect cube.

Solution: (iii) 1331

Taking L.C.M.

$∵ 1331=11 \times 11 \times 11=(11)^{3}$

∴1331 is a perfect cube.

Solution: (iv) 24000

$\begin{array}{c}{∵ 24000=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times} \\ {5 \times 5}\end{array}$

$=(2)^{3} \times(2)^{3} \times(5)^{3} \times 3$

∴24000 is not a perfect cube.

Solution: (v) 1728

Taking L.C.M.

$∵ 1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

$=(2)^{3} \times(2)^{3} \times(3)^{3}$

∴1728 is a perfect cube.

Solution: (vi) 1938

Taking L.C.M.

$1938=2 \times 3 \times 17 \times 19$

1938 is not a perfect cube.

Question 3.

Find the cubes of:

(i) 2.1

(ii) 0.4

(iii) 1.6

(iv) 2.5

(v) 0.12

(vi) 0.02

Solution: (i) 2.1

$2.1=(2.1)^{3}=\left(\frac{21}{10}\right)^{3}=\frac{21 \times 21 \times 21}{10 \times 10 \times 10}$ (Splitting the terms)

$=\frac{9261}{1000}=9.261$

Solution: (ii) 0.4

$0.4=(0.4)^{3}=\left(\frac{4}{10}\right)^{3}=\frac{4 \times 4 \times 4}{10 \times 10 \times 10}$ (Splitting the terms)

$=\frac{64}{1000}=0.064$

Solution: (iii) 1.6

$1.6=(1.6)^{3}=\left(\frac{16}{10}\right)^{3}=\frac{16 \times 16 \times 16}{10 \times 10 \times 10}$ (Splitting the terms)

$=\frac{4096}{1000}=4.096$

Solution: (iv) 2.5

$2.5=(2.5)^{3}=\left(\frac{25}{10}\right)^{3}=\frac{25 \times 25 \times 25}{10 \times 10 \times 10}$

$=\frac{15625}{1000}=15.625$

Solution: (v) 0.12

$0.12=(0.12)^{3}=\left(\frac{12}{100}\right)^{3}=\frac{12 \times 12 \times 12}{100 \times 100 \times 100}$

$=\frac{1728}{1000000}=0.001728$

Solution: (vi) 0.02

$0.02=(0.02)^{3}=\left(\frac{2}{100}\right)^{3}=\frac{2 \times 2 \times 2}{100 \times 100 \times 100}$

$=\frac{8}{1000000}=0.000008$

Solution: (vii) 0.8

$0.8=(0.8)^{3}=\left(\frac{8}{10}\right)^{3}=\frac{8 \times 8 \times 8}{10 \times 10 \times 10}=\frac{512}{1000}=0.512$

### Question 4

Find the cubes of:

(i) $\frac{3}{7}$

(ii) $\frac{8}{9}$

(iii) $\frac{10}{13}$

(iv) $1 \frac{2}{7}$

(v) $2 \frac{1}{2}$

Solution: $(\mathrm{i}) \frac{3}{7}$

(i) $(\mathrm{i}) \frac{3}{7}=\left(\frac{3}{7}\right)^{3}=\frac{3 \times 3 \times 3}{7 \times 7 \times 7}=\frac{27}{343}$

Solution: $(\text { ii }) \frac{8}{9}$

$\frac{8}{9}=\left(\frac{8}{9}\right)^{3}=\frac{8 \times 8 \times 8}{9 \times 9 \times 9}=\frac{512}{729}$

Solution: (iii) $\frac{10}{13}$

$\frac{10}{13}=\left(\frac{10}{13}\right)^{3}=\frac{10 \times 10 \times 10}{13 \times 13 \times 13}=\frac{1000}{2197}$

Solution: $(\text { iv }) 1 \frac{2}{7}$

(iv) $1 \frac{2}{7}=\left(1 \frac{2}{7}\right)^{3}=\left(\frac{1 \times 7+2}{7}\right)^{3}=\left(\frac{9}{7}\right)^{3}$

$=\frac{9 \times 9 \times 9}{7 \times 7 \times 7}=\frac{729}{343}=2 \frac{43}{343}$

Solution: $(\mathrm{v}) 2 \frac{1}{2}$

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

$=\frac{5 \times 5 \times 5}{2 \times 2 \times 2}=\frac{125}{8}=15 \frac{5}{8}$

### Question 5.

Find the cubes of:

(i) -3

(ii) -7

(iii) -12

(iv) -18

(v) -25

(vi) -30

(vii) -50

Solution: (i) -3

(i) $-3=(-3)^{3}=-3 \times-3 \times-3$

$=-(3 \times 3 \times 3)=-27$

Solution: (ii) -7

$-7=(-7)^{3}=-7 \times-7 \times-7$

$=-(7 \times 7 \times 7)=-343$

Solution: (iii) -12

$-12=(-12)^{3}=-12 \times-12 \times-12$

$=-(12 \times 12 \times 12)=-1728$

Solution: (iv) -18

$-18=(-18)^{3}=-18 \times-18 \times-18$

$=-(18 \times 18 \times 18)=-5832$

Solution: (v) -25

$-25=(-25)^{3}=-25 \times-25 \times-25$

$=-(25 \times 25 \times 25)=-15625$

Solution: (vi) -30

$-30=(-30)^{3}=-30 x-30 \times-30$

$=-(30 \times 30 \times 30)=-27000$

Solution: (vii) -50

$=-50=(-50)^{3}=-50 \times-50 \times-50$

$=-50 \times 50 \times 50) =-125000$

### Question 6.

Which of the following are cubes of?

(i) An even number

(ii) An odd number

216,729,3375,8000,125,343,4096 and 9261

Solution:

$216=2 \times 2 \times 2 \times 3 \times 3 \times 3$$=(2)^{3} \times(3)^{3}=(6)^{3}$

$729=3 \times 3 \times 3 \times 3 \times 3 \times 3$

$=(3)^{3} \times(3)^{3}=(9)^{3}$

$3375=5 \times 5 \times 5 \times 3 \times 3 \times 3$

$=(5)^{3} \times(3)^{3}=(15)^{3}$

$8000=20 \times 20 \times 20=(20)^{3}$

$125=5 \times 5 \times 5=(5)^{3}$

$343=7 \times 7 \times 7=(7)^{3}$

$\begin{array}{c}{4096=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} \\ { \times 2 \times 2 \times 2}\end{array}$

$=(2)^{3} \times(2)^{3} \times(2)^{3} \times(2)^{3}=(16)^{3}$

(i) Cubes of an even number are 216, 8000, 4096.

(ii) Cubes of an odd number are 729, 3375, 125, 343, 9261

Question 7.

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

Solution:

The prime factor of 1323 are

$=3 \times 3 \times 3 \times 7 \times 7$

$= (3 \times 3 \times 3) \times 7 \times 7$

Clearly, 1323 must be multiplied by 7.

Question 8.

Find the smallest number by which 8768 must be divided so that the quotient is a perfect cube.

Solution:

The prime factor of 8768 are

$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 137$

$=(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times 137$

Clearly, 8768 must be divided by 137

### Question 9.

Find the smallest number by which 27783 be multiplied to get a perfect square number.

Solution:

$=3 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7$

$=(3 \times 3 \times 3) \times(7 \times 7 \times 7) \times 3$

Clearly, 27783 must be multiplied by

$3 \times 3=9$

### Question 10.

With what least number must 8640 be divided so that the quotient is a perfect cube?

Solution:

The prime factors of 8640 are

$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$

$=(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times(3 \times 3 \times 3) \times 5$

Clearly, 8640 must be divided by 5.

### Question 11.

Which is the smallest number that must be multiplied to 77175 to make it a perfect cube?

Solution:

The prime factors of 77175 are

$=3 \times 3 \times 5 \times 5 \times 7 \times 7 \times 7$

$=(7 \times 7 \times 7) \times 3 \times 3 \times 5 \times 5$

Clearly, 77175 must be multiplied by

$3 \times 5=15$

## EXERCISE 4(B)

### Question 1.

Find the cube-roots of:

(i)64

(ii) 343

(iii) 729

(iv) 1728

(v) 9261

(vi) 4096

(vii) 8000

(viii) 3375

Solution: (i)64

$64=\sqrt[3]{64}=(2 \times 2 \times 2) \times(2 \times 2 \times 2)$

$=2 \times 2=4$

Solution: (ii) 343

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

Solution: (iii) 729

$729==\sqrt[3]{729}=(3 \times 3 \times 3) \times(3 \times 3 \times 3)$

$=3 \times 3=9$

Solution: (iv) 1728

$1728==\sqrt[3]{1728}=(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times(3 \times 3 \times 3)$

$=2 \times 2 \times 3=12$

Solution: (v) 9261

$9261==\sqrt[3]{9261}=(3 \times 3 \times 3) \times(7 \times 7 \times 7)$

$=3 \times 7=21$

Solution: (vi) 4096

$4096=\sqrt[3]{4096}=(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times(2 \times 2 \times 2)$

$=2 \times 2 \times 2 \times 2=16$

Solution: (vii) 8000

$8000=\sqrt[3]{8000}=(4 \times 4 \times 4) \times(5 \times 5 \times 5)$

$=4 \times 5=20$

Solution: (viii) 3375

$3375=\sqrt[3]{3375}=(5 \times 5 \times 5) \times(3 \times 3 \times 3)$

$=5 \times 3=15$

### Question 2:

Find the cube-roots of:

(i) $\frac{27}{64}$

(ii) $\frac{125}{216}$

(iii) $\frac{343}{512}$

(iv) $64 \times 729$

(v) $64 \times 27$

(vi) $729 \times 8000$

(vii) $3375 \times 512$

Solution: (i) $\frac{27}{64}$

$\frac{27}{64}=\sqrt[3]{\frac{27}{64}}=\frac{\sqrt{3 \times 3 \times 3}}{\sqrt{4 \times 4 \times 4}}=\frac{3}{4}$

Solution: (ii) $\frac{125}{216}$

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

Solution: (iii) $\frac{343}{512}$

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

Solution: (iv) $64 \times 729$

$64 \times 729=\sqrt[3]{64 \times 729}$

$=\sqrt{4 \times 4 \times 4 \times 9 \times 9 \times 9}=4 \times 9=36$

Solution: (v) $64 \times 27$

$64 \times 27=\sqrt[3]{64 \times 27}$

$=\sqrt{4 \times 4 \times 4 \times 3 \times 3 \times 3}=4 \times 3=12$

Solution: (vi) $729 \times 8000$

$729 \times 8000=\sqrt[3]{729 \times 8000}$

$=\sqrt{9 \times 9 \times 9 \times 20 \times 20 \times 20}$

$=9 \times 20=180$

Solution: (vii) $3375 \times 512$

$3375 \times 512=\sqrt[3]{3375 \times 512}$

$=\sqrt{15 \times 15 \times 15 \times 8 \times 8 \times 8}$

$=15 \times 8=120$

### Question 3.

Find the cube-roots of:

(i) -216

(ii) -512

(iii) -1331

(iv) $\frac{-27}{125}$

(v) $\frac{-64}{343}$

(vi) $\frac{-512}{343}$

(vii) -2197

(viii) -5832

(ix) -2744000

Solution: (i)-216

$-216=\sqrt[3]{-216}=\sqrt{-6 x-6 x-6}=-6$

Solution: (ii) -512

$-512=\sqrt[3]{-512}=\sqrt{-8 x-8 x-8}=-8$

Solution: (iii) -1331

$-1331=\sqrt[3]{-1331}$

$=\sqrt{-11 x-11 x-11}=-11$

Solution: (iv) $\frac{-27}{125}$

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

Solution: (v) $\frac{-64}{343}$

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

Solution: (vi) $\frac{-512}{343}$

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

Solution: (vii) -2197

$-2197=\sqrt[3]{-2197}$

$=\sqrt[3]{-13 x-13 x-13}=-13$

Solution: (viii) -5832

$-5832=\sqrt[3]{-5832}$

$=\sqrt{-2 x-2 x-2 x-3 x-3 x-3 x-3 x-3 x-3}$

$=-2 \times-3 \times-3=-18$

Solution: (ix) -2744000

$=\sqrt{-2 \times-2 \times-2 \times-7 \times-7 \times-7 \times-10 \times-10 \times-10}$

$=-2 \times-7 \times-10=-140$

### Question 4.

Find the cube-roots of:

(i) 2.744

(ii) 9.261

(iii) 0.000027

(iv) -0.512

(v) -15.625

(vi) $-125 \times 1000$

Solution: (i) 2.744

$2.744=\sqrt[3]{\frac{2744}{1000}}$

$=\sqrt[3]{\frac{2 \times 2 \times 2 \times 7 \times 7 \times 7}{10 \times 10 \times 10}}$

$=\frac{2 \times 7}{10}=\frac{14}{10}=1.4$

Solution: (ii) 9.261

$9.261=\sqrt[3]{\frac{9261}{1000}}$

$=\sqrt{\frac{3 \times 3 \times 3 \times 7 \times 7 \times 7}{10 \times 10 \times 10}}=\frac{3 \times 7}{10}=\frac{21}{10}=2.1$

Solution: (iii) 0.000027

$0.000027=\sqrt[3]{\frac{27}{1000000}}$

$=\sqrt[3]{\frac{3 \times 3 \times 3}{100 \times 100 \times 100}}=\frac{3}{100}=0.03$

Solution: (iv) -0.512

$-0.512=\sqrt[3]{\frac{-512}{1000}}=\sqrt{\frac{-8 x-8 \times-8}{10 \times 10 \times 10}}$

$=\frac{-8}{10}=-0.8$

Solution: (v)= 15.625

$-15.625=\sqrt[3]{\frac{-15625}{1000}}$

$\sqrt{\frac{-(5 \times 5 \times 5) \times(5 \times 5 \times 5)}{10 \times 10 \times 10}}=\frac{-5 \times 5}{10}=\frac{-25}{10}=-2.5$

Solution: (vi) $-125 \times 1000$

$-125 \times 1000=\sqrt{-125 \times 100}$

$=\sqrt{-(5 \times 5 \times 5) \times(10 \times 10 \times 10)}$

$=-5 \times 10=-50$