**Frank Learning Maths Class 8 Chapter 4 Cubes and Cube Roots Exercise 4.2 Solutions**

This post is created to help all the CBSE Class 8 students for the Solutions of Frank Learning Maths Class 8 Mathematics Book, Chapter 4 Cubes and Cube Roots Exercise 4.2. Here students can easily find step by step solutions of all the problems for Cubes and Cube Roots Exercise 4.2. Step by Step proper solutions for every problems. All the problem are solved with easily understandable methods with proper guidance so that all the students can understand easily.

**Chapter 4 – Cubes and Cube Roots**

**Cubes and Cube Roots – Exercise 4.2 all Questions Solution**

**(1) Evaluate **

**Solution : **

**(a)** Given, ∛0.13 × 0.13 × 0.13 × 65 × 65 × 65

= 0.13/100 × 65

= 8.75

**(b)** Given, ∛ -64 + ∛ 0.027/1000

= – 8 + 3/10

= -8 + 3

= – 8.3

**(2) Find the cube root of the following by finding the units and the tens digits
**

**Solution : **

**(a)** 6859

The unit digit 9

Tens digit 1

∴ The cube root of 6859 = 19

**(b)** 24,389

The unit digit = 9

tens digit = 2

∴ The cube root = 29

**(c)** 9261

Unit digit = 1

Tens digit = 2

∴ The cube root = 21

**(d)** 15625

Unit digit = 5

Tens digit = 2

∴ The cube root = 25

**(3) Find the cube root of the following by prime factorisation**

**Solution : **

**(a)** -2197

∴ 3√-2197

= -3√13 × 13 × 13

= -13

**(b)** -5832

∴ – 3√5832

= – 3√2 × 2 × 2 × 9 × 9 × 9

= – (2 × 9)

= – 18

**(c)** 21952

∴ 3√21952

= 3√4 × 4 × 4 × 7 × 7 × 7

= 4 × 7

= 28

**(d)** 13824

∴ 3√13824

= 3√4 × 4 × 4 × 6 × 6 × 6

= 4 × 6

= 24

**(4) Find the cube root of the following rational and decimals. **

**Solution : **

**(a)** 729/1728

= ∛729/1728

= ∛9 × 9 × 9 /12 × 12 × 12

= 9/12

**(b)** -343/2197

= ∛-343/2197

= ∛(-7) × (-7) /13 × 13 × 13

= -7/13

**(c)** 0.004096

∛0.004096/1000000

= ∛16 × 16 × 16 /100 × 100 × 100

= 16/100

= .16

**(d)** -9.261

= ∛- 9261/100

= ∛-21 × (-21) × (-21)/10 × 10 × 10

= -21/10

= – 2.1

**(5) Find the cube root of the following**

**Solution : **

**(a)** 216 × 343

∴ ∛216 × 343

= ∛6 × 6 × 6 × 7 × 7 × 7

= 6 × 7

= 42

Therefore, the cube root of 216 × 343 is 42

**(b)** 144 × 96

∴ ∛144 × 96

= ∛12 × 12 × 4 × 2 × 12

= 12 × 12

= 24

Hence, the cube root of 144 × 96 is 24

**(c)** 250 × 28 × 49

∴ ∛250 × 28 × 49

= ∛5 × 5 × 5 × 2 × 2 × 2 × 7 × 7 × 7

= 5 × 2 × 7

= 70

Thus, the cube root of 250 × 28 × 49 is 70.

**(d)** -216 × 729

∴ ∛-216 × 729

= ∛- 6 × 6 × 6 × 9 × 9 × 9

= – (6 × 9)

= – 54

Therefore, the cube root of -216 × 729 is -54

**(6) Show that**

**Solution : **

**(a)** As per the question,

∛125 × 216 = ∛125 × ∛216

∴ L.H.S = 3√125 × 216

= 3√5 × 5 × 5 × 6 × 6 × 6

= 5 × 6

= 30

∴ R.H.S = 3√125 × 3√216

= 5 × 6

= 30

∴ L.H.S = R.H.S…(Proved)

**(b)** Given, ∛- 125 × 216 = ∛- 125 × ∛216

∴ L.H.S, ∛- 125 × 216

= ∛- 27000

= – 30

R.H.S, ∛- 125 × ∛216

= – 5 × 6

= – 30

∴ L.H.S = R.H.S … [Proved]

**Question no – (7) **

**Solution : **

Given, 2460375 = 3375 × 729

∴ ∛2460375 = ∛33754 × 729

= ∛15 × 15 × 15 × 9 × 9 × 9

= 15 × 9

= 135

Therefore, the cube root of 24,60,375; 2,03,46,417 and 1, 65,81,375 is 135.

**Question no – (8) **

**Solution : **

**(a)** 196

∴ 196 = 2 × 2 × 7 × 7

∴ Multiplied number be = 14

**(b)** 3584

∴ 3584 = 4 × 4 × 4 × 2 × 2 × 2 × 7

∴ Multiplied number = 49

**(c)** 4116

**∴** 4116 = 4 × 3 × 7 × 7 × 7

∴ Multiplied number be 12

**(d)** 1275

∴ 1275 = 5 × 5 × 17 × 3

∴ Multiplied number be,

= 5 × 17 × 17 × 9/3.005

**Question no – (9) **

**Solution : **

**(a)** 725

725 = 5 × 5 × 29

∴ For perfect cube we should divided the number by

= 5 × 5 × 29

= 725

∴ 3√1 = 1

**(b)** 550

∴ For divided 5 × 5 × 2 × 11

= 550

∴ 3√1 = 1

**(c)** 1375

∴ Divided = 5 × 5 × 3 × 17

= 1375

∴ 3√1 = 1

**(d)** 1824

1824 = 2 × 2 × 2 × 2 × 2 × 3 × 19

∴ For perfect cube divided by,

= 2 × 2 × 3 × 19

= 228

∴ 3√8 = 2

**Question no – (10) **

**Solution : **

In the given question we get,

Cube volume is = 729 cm^{3}

length of the edge of a cube = ?

Step by Step Solution :

Given volume of the cube is 729 cm^{3}

So now,

Therefore, the length of the edge of the cube will be 9 cm.

**Question no – (11) **

**Solution : **

(2x)^{3} + (3x)^{3} + (4x)^{3} = 33957

8x^{3} + 27x^{3} + 64x^{3} = 33957

99x^{3} = 33957

x^{3} = 33957/99

x = 3√343 = 7

Therefore, the required number are 14, 21 and 28

**Question no – (12) **

**Solution : **

Let the length of cube is = a

∴ 6a^{2} = 726

∴ a^{2} = 121

∴ a = 11

**∴ Volume, **

= 11 × 11 × 11

= 1331 m^{3}

Therefore, the Volume will be 1331 m^{3}

**Question no – (13) Simplify**

**Solution : **

**(a)** ∛64 + ∛.512 – ∛0.125

= 8 + 3√512/100 – 3√125/100

= 8 + 8/10 – 5/10

= 8 + 3/10

= 8 + .3

= 8.3…(Simplified)

**(b)** ∛729/216 × 2

= 9/16 × 2

= 3…(Simplified)

**(c)** ∛0.008/0.125 + √0.16/0.09 – 2

= 2/5 + 4/3 – 2

= 6 + 20 – 30/15

= – 4/15…(Simplified)