## 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.

### Download Exclusively Curated Chapter Notes for Class 8 Maths Chapter – 7 Cubes and Cube Roots

### Download Most Important Questions for Class 8 Maths Chapter – 7 Cubes and Cube Roots

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 1^{3} = 1 and 2^{3} = 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 2^{3} = 8 and 3^{3} = 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 3^{3} = 27 and 4^{3} = 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.

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