NCERT Solutions Class 8 Maths Chapter 6 Cubes and Cube Root Exercise 6.2

In Exercise 6.2 of Chapter 6, you will learn how cube root works and how to find them using prime factorisation. The questions are basic, easy, and will teach you about cube roots step by step. These are useful concepts not only for your CBSE examinations, but also for competitions like Math Olympiads as questions related to numbers often appear. 

The NCERT Solutions in this exercise has been taken from the new updated NCERT Maths syllabus pattern for Class 8, as per the official NCERT text book. The solutions are in a simple step by step manner allowing you to understand the concepts of cube roots and use the methods properly. Through regular practice of exercise 6.2, you will be able to approach cube roots confidently and more quickly, thus gaining more confidence in Maths.

1.0Download NCERT Solutions Class 8 Maths Chapter 6 Cubes and Cube Root Exercise 6.2: Free PDF

Understand how to easily solve cube root questions with NCERT Solutions for Class 8 Maths Chapter 6 here. The free pdf has all the answers from Exercise 6.2 in easy steps. Click the link below to download for free.

2.0Key Concepts in Exercise 6.2 of Class 8 Maths Chapter 6

This exercise focuses on finding cube roots of numbers through a clear method. Key concepts include:

• Understanding cube roots
• Cube roots of perfect cube numbers
• Prime factorisation method to find cube roots
• Identifying if a number is a perfect cube
• Using basic number properties to find cube roots

3.0NCERT Class 8 Maths Chapter 6: Other Exercises

4.0NCERT Class 8 Maths Chapter 6 Exercise 6.2: Detailed Solutions

• 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 Sol. (i) Resolving 64 into prime factors, we get

$64=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$ $\therefore \sqrt[3]{64}=(2\times 2)=4$ (ii) Resolving 512 into prime factors, we get

$512=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$ $\therefore \sqrt[3]{512}=(2\times 2\times 2)=8$ (iii) Resolving 10648 into prime factors, we get

$10648=\underline{2\times 2\times 2}\times \underline{11\times 11\times 11}$ $\therefore \sqrt[3]{10648}=(2\times 11)=22$ (iv) Resolving 27000 into prime factors, we get

$27000=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\times \underline{5\times 5\times 5}$ $\therefore \sqrt[3]{27000}=(2\times 3\times 5)=30$ (v) Resolving 15625 into prime factors, we get

$15625=\underline{5\times 5\times 5}\times \underline{5\times 5\times 5}$ $\therefore \sqrt[3]{15625}=(5\times 5)=25$ (vi) Resolving 13824 into prime factors, we get

$13824=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times$ $\underline{3\times 3\times 3}$ $\therefore \sqrt[3]{13824}=(2\times 2\times 2\times 3)=24$ (vii) Resolving 110592 into prime factors, we get

$110592=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2\times \underline{2\times 2\times 2}}$ $\times \underline{2\times 2\times 2\times \underline{3\times 3\times 3}}$ $\therefore \sqrt[3]{110592}=(2\times 2\times 2\times 2\times 3)=48$ (viii) Resolving 46656 into prime factors, we get

$46656=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\times$ $3\times 3\times 3$ $\therefore \sqrt[3]{46656}=(2\times 2\times 3\times 3)=36$ (ix) Resolving 175616 into prime factors, we get

$175616=\underline{2\times 2\times 2\times 2\times 2\times 2\times \underline{2\times 2\times 2}}$ $\times \underline{7\times 7\times 7}$ $\therefore \sqrt[3]{175616}=(2\times 2\times 2\times 7)=56$ (x) Resolving 91125 into prime factors, we get

$91125=\underline{3\times 3\times 3}\times \underline{3\times 3\times 3\times 5\times 5\times 5}$ $\therefore \sqrt[3]{91125}=(3\times 3\times 5)=45$

• State true or false (i) Cube of any odd number is even. (ii) A perfect cube does not end with two zeros. (iii) If 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 two digit number may have seven or more digits. (vii) The cube of a single digit number may be a single digit number. Sol. (i) False, as $3^3=27$ is odd (ii) True, as $10^3=1000$ (iii) False, as $15^2=225$ and $15^3=3375$ (iv) False, as $8=2^3,1728=12^3$ etc. (v) False, as $10^3=1000$ (vi) False, as $99^3=970299$ (vii) True as $2^3=8$ and 8 is a single digit number.
• 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 $4913,12167,32768$. Sol. (i) 1331 (1) Form group of three starting from the right most digit of 1331 . 1 331. In this case one group i.e., 331 has three digits whereas 1 has only one digit. (2) Take 331 . The digit 1 is at one's place. We take the one's place of the required cube root as 1 . (3) Take the other group, i.e. 1. Cube of 1 is 1 . So, take 1 as ten's place of the cube root of 1331 . Thus, $\sqrt[3]{1331}=11$ (ii) 4913 (1) Form group of three starting from the right most digit of 4913 . 4 913. In this case one group i.e., 913 has three digits whereas 4 has only one digit. (2) Take 913. The digit 3 is at one's place. We take the one's place of the required cube root as 7 . (3) Take the other group, i.e., 4. Cube of 1 is 1 and cube of 2 is 8.4 lies between $1\&8$. The smaller number among 1 and 2 is 1 . So, take 1 as ten's place of the cube root of 4913. Thus, $\sqrt[3]{4913}=17$ (iii) 12167 (1) Form group of three starting from the right most digit of 12167 . 12167 . In this case one group i.e., 167 has three digits whereas 12 has only two digits. (2) Take 167. The digit 7 is at its one's place. We take the one's place of the required cube root as 3 . (3) Take the other group, i.e., 12. Cube of 2 is 8 and cube of 3 is 27.12 lies between 8 & 27. The smaller number among 2 and 3 is 2 . So, take 2 as ten's place of the cube root of 12167 . Thus, $\sqrt[3]{12167}=23$ (iv) 32768 (1) Form group of three starting from the right most digit of 32768 . 32 768. In this case one group i.e., 768 has three digits whereas 32 has only two digits. (2) Take 768. The digit 8 is at its one's place. We take the one's place of the required cube root as 2 . (3) Take the other group, i.e., 32. Cube of 3 is 27 and cube of 4 is 64 . 32 lies between 27 & 64. The smaller number among 3 and 4 is 3 . So, take 3 as ten's place of the cube root of 32768 . Thus $\sqrt[3]{32768}=32$

5.0Key Features and Benefits of Class 8 Maths Chapter 6 Exercise 6.2

• Introduces cube roots with examples and step-wise explanations.
• The exercise follows the NCERT Class 8 Maths syllabus and pattern.
• Here, the prime factor method is used, which helps to make cube root calculation simple and logical.
• Regular practice builds confidence solving cube root related questions in the exams.
• Helps improve logical thinking skills and also sharpen problem solving ability as students go into higher classes.