NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots( New Syllabus)

The NCERT Solutions for Class 8 Maths Chapter 6 given in Study Circle contain solutions to all the questions covered under Chapter 6 Cubes and Cube Roots of Class 8 NCERT textbook.

By following the concepts applied in NCERT Solutions for Class 8 Maths Chapter 6, students will easily solve all their questions related to cube and finding cube roots. These answers have been prepared by Study Circle as per the latest CBSE syllabus.

NCERT Solutions for Class 8 Maths Chapter 6

Exercise 6.1

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

(i) 216

Solution:

By resolving 216 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

(ii) 128

Solution:

By resolving 128 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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, and we are left with one factor: 2.

∴ 128 is not a perfect cube.

(iii) 1000

Solution:

By resolving 1000 into prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the cube of 10.

(iv) 100

Solution:

By resolving 100 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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,

NCERT Solutions for Class 8 Maths Chapter 6

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 the cube of 36.

Question 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 a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

(ii) 256

Solution:

By resolving 256 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

(iii) 72

Solution:

By resolving 72 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

(iv) 675

Solution:

By resolving 675 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

(v) 100

Solution:

By resolving 100 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

Question 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 a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

(ii) 128

Solution:

By resolving 128 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

(iii) 135

Solution:

By resolving 135 into prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

(iv) 192

Solution:

By resolving 192 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

(v) 704

Solution:

By resolving 704 into a prime factor,

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube.

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

Solution:

Given the sides of the cube are 5 cm, 2 cm and 5 cm.

∴ Volume of cube = 5×2×5 = 50

NCERT Solutions for Class 8 Maths Chapter 6

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 the perfect cube. Hence, 20 cuboids are needed.

NCERT Solutions for Class 8 Maths Chapter 6

Exercise 6.2

Question 1. Find the cube root of each of the following numbers by the 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 the 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 the 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 the 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 the 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 the 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 the 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 the 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 the 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 the 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 the cube root of 91125.

Question 2. State true or false.

(i) Cube of any odd number is even.

Solution:

False

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

Solution:

True

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

Solution:

False

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

Solution:

False

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

Solution:

False

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

Solution:

False

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

Solution:

True.

Question 3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

Solution:

(i) By grouping the digits, we get 1 and 331

We know that since the unit digit of the cube is 1, the unit digit of the cube root is 1.

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

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

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

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

∴ ∛1331 = 11

(ii) By grouping the digits, we get 4 and 913

We know that since the unit digit of the cube is 3, the unit digit of the cube root is 7.

∴ we get 7 as the unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8 , 1 > 4 > 8

Thus, 1 is taken as the tens digit of the cube root.

∴ ∛4913 = 17

(iii) By grouping the digits, we get 12 and 167.

We know that since the unit digit of the cube is 7, the unit digit of the cube root is 3.

∴ 3 is the unit digit of the cube root of 12167 We know 23 = 8 and 33 = 27 , 8 > 12 > 27

Thus, 2 is taken as the tens digit of the cube root.

∴ ∛12167= 23

(iv) By grouping the digits, we get 32 and 768.

We know that since the unit digit of the cube is 8, the unit digit of the cube root is 2.

∴ 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64 , 27 > 32 > 64

Thus, 3 is taken as the tens digit of the cube root.

∴ ∛32768= 32

 

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