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

## NCERT Solutions Class 8 Maths Chapter 7 – Free PDF Download

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 BYJU’S, as per the latest CBSE Syllabus.

As Class 8 is a critical stage in their academic career, these NCERT Solutions provide extensive knowledge about the concepts covered. BYJU’S expert team has solved the questions in NCERT Solutions Chapter 7 in a step-by-step format, which helps the students strengthen their concepts. The concepts discussed in this chapter include the 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.

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

### Access Answers of Maths NCERT Class 8 Chapter 7 – Cubes and Cube Roots

Exercise 7.1 Page: 114

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

(i) 216

Solution:

By resolving 216 into a 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 the cube of 6.

(ii) 128

Solution:

By resolving 128 into a 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, and we are left with one factor: 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 the cube of 10.

(iv) 100

Solution:

By resolving 100 into a 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 the 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 a 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 the perfect cube.

(ii) 256

Solution:

By resolving 256 into a 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 the perfect cube.

(iii) 72

Solution:

By resolving 72 into a 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 the perfect cube.

(iv) 675

Solution:

By resolving 675 into a 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 the perfect cube.

(v) 100

Solution:

By resolving 100 into a 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 the 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 a 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 the perfect cube.

(ii) 128

Solution:

By resolving 128 into a 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 the 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 the perfect cube.

(iv) 192

Solution:

By resolving 192 into a 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 the perfect cube.

(v) 704

Solution:

By resolving 704 into a 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 the perfect cube. 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 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. ## Exercise 7.2 Page: 116

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. 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

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

 Also Access NCERT Exemplar for Class 8 Maths Chapter 7 CBSE Notes for Class 8 Maths Chapter 7

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

The NCERT solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots provided by BYJU’S contain 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. Let’s take a glance at what the chapter discusses on “Cubes and Cube Roots”.

• Numbers like 1729, 4104, and 13832 are known as Hardy–Ramanujan Numbers. They can be expressed as the 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 Question)

### 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 the prime factorisation method. Moreover, 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.

Disclaimer:

Dropped Topics – 7.3.2 Cube root of a cube number.

## Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 7

Q1

### What kind of questions are there in NCERT Solutions for Class 8 Maths Chapter 7?

Chapter 7 of NCERT Solutions for Class 8 Maths has multiple choice questions, descriptive type questions, long answer type questions, short answer type questions, fill in the blanks and daily life examples. By the end of this chapter, students can increase their problem-solving skills and time-management skills. This helps in procuring high marks in their finals.
Q2

### Are NCERT Solutions for Class 8 Maths Chapter 7 enough to attend all the questions that come in the board exam?

Yes, it is enough to solve all the questions that come in the board exam of NCERT Solutions for Class 8 Maths Chapter 7. Practising this chapter can make them learn the concepts flawlessly. These questions have been devised as per the NCERT syllabus and the guidelines. This makes the students score good marks in the finals.
Q3

### Is it necessary to learn all the topics provided in NCERT Solutions for Class 8 Maths Chapter 7?

Yes, it is compulsory to learn all the topics provided in NCERT Solutions for Class 8 Maths Chapter 7 to score high marks in Class 8 board marks. These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook. They also focus on cracking the solutions of Maths in such a way that it is easy for the students to understand.