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Our NCERT Solutions Class 8 Math Chapter 7 Pair of Cubes and Cube Roots will assist you in understanding and resolving all NCERT question-answer concerns. We've covered every topic and answered every in-text and exercise question. For one-stop solutions for all concepts of Cubes and Cube Roots, please see our solution. Our NCERT solutions are created by a subject matter specialist who understands all of the concepts in this field. Experts have used various methods to make these solutions easier to understand for users, such as diagrams, different equations, graphs, shortcuts, tips, and tricks.

We have provided NCERT Solutions Class 8 with all the related topics that deal with the students' queries in this chapter of Cubes and Cube Roots. The following are some of the chapter's key points:

  • Perfect cube
  • Adding consecutive odd numbers
  • Cubes and their prime factors
  • Smallest multiple that is a perfect cube
  • Cube Roots

Our NCERT Solutions Class 8 Maths Chapter 7 Cubes and Cube Roots will assist students in easily comprehending the concepts. Our NCERT solution is ideal for last-minute examination preparation. To get good grades, read through all of the solutions provided here.

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NCERT Solutions Class 8 Maths Chapter 7 Cubes and Cube Roots

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

(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656

Solution:

(i) Prime factorisation of 216 is:

216 = 2 × 2 × 2 × 3 × 3 × 3

In the above factorisation, 2 and 3 have formed a group of three.

Thus, 216 is a perfect cube.

(ii) Prime factorisation of 128 is:

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, 2 is left without making a group of three.

Thus 128 is not a perfect cube.

(iii) Prime factorisation of 1000, is:

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, no number is left for making a group of three.

Thus, 1000 is a perfect cube.

(iv) Prime factorisation of 100, is:

100 = 2 × 2 × 5 × 5

Here 2 and 5 have not formed a group of three.

Thus, 100 is not a perfect cube.

(v) Prime factorisation of 46656 is:

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here 2 and 3 have formed the groups of three.

Thus, 46656 is a perfect cube.

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

(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100

Solution:

(i) Prime factorisation of 243, is:

243 = 3 × 3 × 3 × 3 × 3 = 33 × 3 × 3

Here, number 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3

Thus, the required smallest number to be multiplied is 3.

(ii) Prime factorisation of 256, is:

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 2 × 2

Here, a number 2 is needed to make 2 × 2 a group of three, i.e., 2 × 2 × 2

Thus, the required smallest number to be multiplied is 2.

(iii) Prime factorisation of 72, is:

72 = 2 × 2 × 2 × 3 × 3 = 23 × 3 × 3

Here, a number 3 is required to make 3 × 3 a group of three, i.e. 3 × 3 × 3

Thus, the required smallest number to be multiplied is 3.

(iv) Prime factorisation of 675, is:

675 = 3 × 3 × 3 × 5 × 5 = 33 × 5 × 5

Here, a number 5 is required to make 5 × 5 a group of three to make it a perfect cube, i.e. 5 × 5 × 5

Thus, the required smallest number is 5.

(v) Prime factorisation of 100, is:

100 = 2 × 2 × 5 × 5

Here, number 2 and 5 are needed to multiplied 2 × 2 × 5 × 5 to make it a perfect cube, i.e., 2 × 2 × 2 × 5 × 5 × 5

Thus, the required smallest number to be multiplied is 2 × 5 = 10.

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

(i) 81
(ii) 128
(iii) 135
(iv) 92
(v) 704

Solution:

(i) Prime factorisation of 81, is:

81 = 3 × 3 × 3 × 3 = 33 × 3

Here, a number 3 is the number by which 81 is divided to make it a perfect cube,

i.e., 81 ÷ 3 = 27 which is a perfect cube.

Thus, the required smallest number to be divided is 3.

(ii) Prime factorisation of 128, is:

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 2

Here, a number 2 is the smallest number by which 128 is divided to make it a perfect cube,

i.e., 128 ÷ 2 = 64 which is a perfect cube.

Thus, 2 is the required smallest number.

(iii) Prime factorisation of 135 is:

135 = 3 × 3 × 3 × 5 = 33 × 5

Here, 5 is the smallest number by which 135 is divided to make a perfect cube,

i.e., 135 ÷ 5 = 27 which is a perfect cube.

Thus, 5 is the required smallest number.

(iv) Prime factorisation of 192 is:

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 23 × 23 × 3

Here, 3 is the smallest number by which 192 is divided to make it a perfect cube,

i.e., 192 ÷ 3 = 64 which is a perfect cube.

Thus, 3 is the required smallest number.

(v) Prime factorisation of 704 is:

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 23 × 23 × 11

Here, 11 is the smallest number by which 704 is divided to make it a perfect cube,

i.e., 704 ÷ 11 = 64 which is a perfect cube.

Thus, 11 is the required smallest number.

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

Solution:

The sides of the cuboid are given as 5 cm, 2 cm and 5 cm.

Volume of the cuboid = 5 cm × 2 cm × 5 cm = 50 cm3

For the prime factorisation of 50, we have

50 = 2 × 5 × 5

To make it a perfect cube, we must have

2 × 2 × 2 × 5 × 5 × 5

= 20 × (2 × 5 × 5)

= 20 × volume of the given cuboid

Thus, the required number of cuboids = 20.

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

(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125

 Solution: 

(i) 64

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

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

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

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

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

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

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

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

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

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.

6. State True or False.

(i) Cube of an odd number is even.

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

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

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

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

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

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

Solution:

(i) False – Cube of any odd number is always odd, e.g., (7)3 = 343

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

(iii) True – If a square of a number ends with 5, then its cube ends with 25, e.g., (5)2 = 25 and (5)3 = 625

(iv) False – (12)3 = 1728 (ends with 8)

(v) False – (10)3 = 1000 (4-digit number)

(vi) False – (99)3 = 970299 (6-digit number)

(vii) True – (2)3 = 8 (1-digit number)

7. 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:

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

We know that, since, the unit digit of cube is 1, the unit digit of cube root 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’s digit of our cube root is taken as the unit place of 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 cube is 3, the unit digit of cube root is 7.

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

Thus, 1 is taken as ten digit of cube root.

∴ ∛4913 = 17

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

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

∴ ∛12167 = 23

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

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

∴ ∛32768 = 32

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