NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

The NCERT Solutions provided here comprise a comprehensive analysis of all the questions that fall under Chapter 7 Cubes and Cube roots of Class 8 NCERT Textbook. Following the notions applied in NCERT Solutions for Class 8, students will be capable of clearing all their doubts associated with obtaining the Cubes and Cube roots. These answers are devised by subject experts at CoolGyan’S, as per the latest CBSE Syllabus.

Class 8 being a critical stage in their academic career, these NCERT Solutions provide extensive knowledge about the concepts covered. CoolGyan’S expert team has solved the questions from Chapter 7 of NCERT Class 8 textbook in a step by step format in detail, which helps the students strengthen their concepts. The concepts discussed in this chapter include cube of a number, finding a cube of a two-digit number by column method, Cubes of Negative Integers, Cubes of Rational Numbers, Cube root of a Natural Number, Cube root of a negative perfect cube, Cube root of the product of integers, finding cube roots using cube root tables.

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Exercise 7.1 Page: 114

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 cube.

(ii) 256 Solution:

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 cube.

(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 cube.

(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 cube.

(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 cube.

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 cube.

(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 cube.

(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 cube.

(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 cube.

(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 cube.

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

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 cube. Hence, 20 cuboid is needed.

Exercise 7.2 Page: 116

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.

Solution:

False

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

Solution:

True

(iii) If 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

3. 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

NCERT Solutions for Class 8 Maths Chapter 7- Cubes and Cube roots

The NCERT solutions for Class 8 Maths Chapter 7 Cubes and Cube roots provided by CoolGyan’S contains the answers for all the questions present in the chapter. The chapter contains two exercises, in which, Exercise 7.1 deals with Cubes and Exercise 7.2 deals with Cube roots. Here is a glance of what the chapter discusses on “Cubes and Cube Roots”.

• Numbers like 1729, 4104, 13832, are known as Hardy – Ramanujan Numbers. They can be expressed as sum of two cubes in two different ways.
• Numbers obtained when a number is multiplied by itself three times are known as cube numbers.
• If in the prime factorisation of any number each factor appears three times, then the number is a perfect cube.

The main topics covered in this chapter include: 7.1 Introduction 7.2 Cubes 7.2.1 Some Interesting Patterns 7.2.2 Smallest multiple that is a perfect cube 7.3 Cube Roots 7.3.1 Cube root through prime factorisation method 7.3.2 Cube root of a cube number
Exercise 7.1 Solutions 4 Questions (4 Short Answer Questions)
Exercise 7.2 Solutions 3 Questions (2 Long Answer Questions,1 Short Answer Questions)

NCERT Solutions for Class 8 Maths Chapter 7- Cubes and Cube roots

The seventh chapter of Class 8 Maths helps the students in finding the cubes and cube roots of different numbers, understanding the difference between cubes and cube roots, and also to have fun with some interesting patterns. The chapter also lets the students understand the process of finding out the cubes and cube roots of a number using prime factorisation method. The chapter also explains the method of finding the cube root of a cube number. Learning the chapter “Cubes and Cube Roots” enables the students to:

• Find out the Cubes and cubes roots for numbers containing at most 3 digits
• Estimating cube roots and cube roots.
• Learning the process of moving nearer to the required number.