Cubes And Cube Roots Chapter 7 Class 8 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 8 Maths Chapter 7′ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 8 Maths Chapter 7 Exercise Solution’ using the download button.

NCERT Class 8 Maths Textbook Chapter 7 With Answer Book PDF Free Download

NCERT Class 8 Maths Textbook Chapter 7 With Answer Book PDF Free Download

Chapter 7:Cubes And Cube Roots

7.1 Introduction

This is a story about one of India’s great mathematical geniuses, S. Ramanujan. Once another famous mathematician Prof. G.H. Hardy came to visit him in a taxi whose number was 1729.

While talking to Ramanujan, Hardy described this number as “a dull number”. Ramanujan quickly pointed out that 1729 was indeed interesting. He said it is the smallest number that can be expressed as a sum of two cubes in two different ways:

7.2 Cubes

You know that the word ‘cube’ is used in geometry. A cube is a solid figure which has all its sides equal.

How many cubes of side 1 cm will make a cube of side 2 cm?
How many cubes of side 1 cm will make a cube of side 3 cm?
Consider the numbers 1, 8, 27, …

These are called perfect cubes or cube numbers. Can you say why they are named so? Each of them is obtained when a number is multiplied by taking it three times.

AuthorNCERT
Language English
No. of Pages8
PDF Size681 KB
CategoryMathematics
Source/ Creditsncert.nic.in

NCERT Solutions Class 8 Maths Chapter 7 Cubes And Cube Roots

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

(i) 216

Solution:

By resolving 216 into prime factor,

NCERT Solution For Class 8 Maths Chapter 7 Image 1

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 triples of equal factors,

∴ 216 = (2×3) = 6

Hence, 216 is a cube of 6.

(ii) 128

Solution:

By resolving 128 into prime factor,

NCERT Solution For Class 8 Maths Chapter 7 Image 2

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 triples of equal factors, we are left off with one factor 2.

∴ 128 is not a perfect cube.

(iii) 1000

Solution:

By resolving 1000 into prime factor,

NCERT Solution For Class 8 Maths Chapter 7 Image 3

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 triples of equal factors,

∴ 1000 = (2×5) = 10

Hence, 1000 is a cube of 10.

(iv) 100

Solution:

By resolving 100 into prime factor,

NCERT Solution For Class 8 Maths Chapter 7 Image 4

100 = 2×2×5×5

Here, 100 cannot be grouped into triples of equal factors.

∴ 100 is not a perfect cube.

NCERT Class 8 Maths Textbook Chapter 7 With Answer Book PDF Free Download

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