#### NCERT Solutions for Class 8th: Ch 7 Cube and Cube Roots Maths

Page No: 114**Exercise 7.1**

1. Which of the following numbers are not perfect cubes?

(i) 216

**Solution**

By resolving 216 into prime factor,

216 = 2×2×2×3×3×3

By grouping the factors in triplets of equal factors,

216 = (2×2×2)×(3×3×3)

Here, 216 can be grouped into triplets of equal factors,

∴ 216 = (2×3) = 6

Hence, 216 is cube of 6.

(ii) 128

**Solution**

By resolving 128 into prime factor,

128 = 2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors,

128 = (2×2×2)×(2×2×2)×2

Here, 128 cannot be grouped into triplets of equal factors, we are left of with one factors 2 .

∴ 128 is not a perfect cube.

(iii) 1000

**Solution**

By resolving 1000 into prime factor,

1000 = 2×2×2×5×5×5

By grouping the factors in triplets of equal factors,

1000 = (2×2×2)×(5×5×5)

Here, 1000 can be grouped into triplets of equal factors,

∴ 1000 = (2×5) = 10

Hence, 1000 is cube of 10.

(iv) 100

**Solution**

By resolving 100 into prime factor,

100 = 2×2×5×5

Here, 100 cannot be grouped into triplets of equal factors.

∴ 100 is not a perfect cube.

(v) 46656

**Solution**

By resolving 46656 into prime factor,

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

By grouping the factors in triplets of equal factors,

46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

Here, 46656 can be grouped into triplets of equal factors,

∴ 46656 = (2×2×3×3) = 36

Hence, 46656 is cube of 36.

2.Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

**Solution**

By resolving 243 into prime factor,

243 = 3×3×3×3×3

By grouping the factors in triplets of equal factors,

243 = (3×3×3)×3×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will multiply 243 by 3 to get perfect square.

(ii) 256

**Soliution**

By resolving 256 into prime factor,

256 = 2×2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors,

256 = (2×2×2)×(2×2×2)×2×2

Here, 2 cannot be grouped into triplets of equal factors.

∴ We will multiply 256 by 2 to get perfect square.

(iii) 72

**Solution**

By resolving 72 into prime factor,

72 = 2×2×2×3×3

By grouping the factors in triplets of equal factors,

72 = (2×2×2)×3×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will multiply 72 by 3 to get perfect square.

(iv) 675

**Solution**

By resolving 675 into prime factor,

675 = 3×3×3×5×5

By grouping the factors in triplets of equal factors,

675 = (3×3×3)×5×5

Here, 5 cannot be grouped into triplets of equal factors.

∴ We will multiply 675 by 5 to get perfect square.

(v) 100

**Solution**

By resolving 100 into prime factor,

100 = 2×2×5×5

Here, 2 and 5 cannot be grouped into triplets of equal factors.

∴ We will multiply 100 by (2×5) 10 to get perfect square.

3.Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

**Solution**

By resolving 81 into prime factor,

81 = 3×3×3×3

By grouping the factors in triplets of equal factors,

81 = (3×3×3)×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will divide 81 by 3 to get perfect square.

(ii) 128

**Solution**

By resolving 128 into prime factor,

128 = 2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors,

128 = (2×2×2)×(2×2×2)×2

Here, 2 cannot be grouped into triplets of equal factors.

∴ We will divide 128 by 2 to get perfect square.

(iii) 135

**Solution**

By resolving 135 into prime factor,

135 = 3×3×3×5

By grouping the factors in triplets of equal factors,

135 = (3×3×3)×5

Here, 5 cannot be grouped into triplets of equal factors.

∴ We will divide 135 by 5 to get perfect square.

(iv) 192

**Solution**

By resolving 192 into prime factor,

192 = 2×2×2×2×2×2×3

By grouping the factors in triplets of equal factors,

192 = (2×2×2)×(2×2×2)×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will divide 192 by 3 to get perfect square.

(v) 704

**Solution**

By resolving 704 into prime factor,

704 = 2×2×2×2×2×2×11

By grouping the factors in triplets of equal factors,

704 = (2×2×2)×(2×2×2)×11

Here, 11 cannot be grouped into triplets of equal factors.

∴ We will divide 704 by 11 to get perfect square.

4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Solution

Given, side of cube is 5 cm, 2 cm and 5 cm.

∴ Volume of cube = 5×2×5 = 50

∴ Volume of cube = 5×2×5 = 50

50 = 2×5×5

Here, 2 , 5 and 5 cannot be grouped into triplets of equal factors.

∴ We will multiply 50 by (2×2×5) 20 to get perfect square.

Hence, 20 cuboid is needed.

Page No. 116

**Exercise 7.2**

1. Find the cube root of each of the following numbers by prime factorisation method

(i) 64

**Solution**

64 = 2×2×2×2×2×2

By grouping the factors in triplets of equal factors,

64 = (2×2×2)×(2×2×2)

Here, 64 can be grouped into triplets of equal factors,

∴ 64 = 2×2 = 4

Hence, 4 is cube root of 64.

(ii) 512

**Solution**

512 = 2×2×2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors,

512 = (2×2×2)×(2×2×2)×(2×2×2)

Here, 512 can be grouped into triplets of equal factors,

∴ 512 = 2×2×2 = 8

Hence, 8 is cube root of 512.

(iii) 10648

**Solution**

10648 = 2×2×2×11×11×11

By grouping the factors in triplets of equal factors,

10648 = (2×2×2)×(11×11×11)

Here, 10648 can be grouped into triplets of equal factors,

∴ 10648 = 2 ×11 = 22

Hence, 22 is cube root of 10648.

(iv) 27000

**Solution**

27000 = 2×2×2×3×3×3×3×5×5×5

By grouping the factors in triplets of equal factors,

27000 = (2×2×2)×(3×3×3)×(5×5×5)

Here, 27000 can be grouped into triplets of equal factors,

∴ 27000 = (2×3×5) = 30

Hence, 30 is cube root of 27000.

(v) 15625

**Solution**

15625 = 5×5×5×5×5×5

By grouping the factors in triplets of equal factors,

15625 = (5×5×5)×(5×5×5)

Here, 15625 can be grouped into triplets of equal factors,

∴ 15625 = (5×5) = 25

Hence, 25 is cube root of 15625.

(vi) 13824

**Solution**

13824 = 2×2×2×2×2×2×2×2×2×3×3×3

By grouping the factors in triplets of equal factors,

13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 13824 can be grouped into triplets of equal factors,

∴ 13824 = (2×2× 2×3) = 24

Hence, 24 is cube root of 13824.

(vii) 110592

**Solution**

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3

By grouping the factors in triplets of equal factors,

110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 110592 can be grouped into triplets of equal factors,

∴ 110592 = (2×2×2×2 × 3) = 48

Hence, 48 is cube root of 110592.

(viii) 46656

**Solution**

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

By grouping the factors in triplets of equal factors,

46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

Here, 46656 can be grouped into triplets of equal factors,

∴ 46656 = (2×2×3×3) = 36

Hence, 36 is cube root of 46656.

(ix) 175616

**Solution**

175616 = 2×2×2×2×2×2×2×2×2×7×7×7

By grouping the factors in triplets of equal factors,

175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)

Here, 175616 can be grouped into triplets of equal factors,

∴ 175616 = (2×2×2×7) = 56

Hence, 56 is cube root of 175616.

(x) 91125

**Solution**

91125 = 3×3×3×3×3×3×3×5×5×5

By grouping the factors in triplets of equal factors,

91125 = (3×3×3)×(3×3×3)×(5×5×5)

Here, 91125 can be grouped into triplets of equal factors,

∴ 91125 = (3×3×5) = 45

Hence, 45 is cube root of 91125.

2.State true or false.

(i) Cube of any odd number is even.

**Ans.**False

(ii) A perfect cube does not end with two zeros.

**Ans.**True

(iii) If square of a number ends with 5, then its cube ends with 25.

**Ans.**False

(iv) There is no perfect cube which ends with 8.

**Ans.**False

(v) The cube of a two digit number may be a three digit number.

**Ans.**False

(vi) The cube of a two digit number may have seven or more digits.

**Ans.**False

(vii) The cube of a single digit number may be a single digit number.

**Ans.**True

4. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

**Solution**

• By grouping the digits, we get 1 and 331

As we know that the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.

∴ We get 1 as unit digit of the cube root of 1331.

The cube of 1 matches with the number of second group.

∴ The ten digit of our cube root is taken as the unit place of smallest number.

As we know that the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.

∴ = 11

• By grouping the digits, we get 4 and 913

As we know that the unit’s digit of the cube of a number having digit as unit’s place 3 is 7.

∴ we get 7 as unit digit of the cube root of 4913.

We know 1

^{3 }= 1 and 2^{3 }= 8 , 1 > 4 > 8.
Thus, 1 is taken as ten digit of cube root.

∴ = 17

• By grouping the digits, we get 12 and 167.

As we know that the unit’s digit of the cube of a number having digit as unit’s place 7 is 3.

∴ we get 3 as unit digit of the cube root of 12167

We know 2

^{3 }= 8 and 3^{3 }= 27 , 8 > 12 > 27.
Thus, 2 is taken as ten digit of cube root.

∴ = 23

• By grouping the digits, we get 32 and 768.

As we know that the unit’s digit of the cube of a number having digit as unit’s place 8 is 2.

∴ we get 2 as unit digit of the cube root of 32768.

We know 3

^{3}= 27 and 4^{3}= 64 , 27 > 32 > 64.
Thus, 3 is taken as ten digit of cube root.

∴ = 32