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Chapter 6: Square Roots and Cube Roots

    Introduction

    Properties of Square Numbers

    Some More Interesting Patterns

    Pythagorean Triplets

    Square roots

    Finding the Square root through Subtraction of successive odd numbers

    Finding the Square root through Prime Factorisation Method

    Finding Square Root by Division Method

    Square roots of Decimals Using Division Method

    Estimating Square Roots of non-perfect Square Numbers

    Cubic Numbers

    Some Interesting Patterns

    Cubes and their Prime Factors

    Cube Roots

    Finding Cube root through Prime Factorisation Method

    Estimating the Cube Root of a Number

    Exercise 6.1

    Exercise 6.2

    Exercise 6.3

    Exercise 6.4

    Exercise 6.5

    What We Have Discussed?

    Easy Level Worksheet Questions

    Moderate Level Worksheet Questions

    Hard Level Worksheet Questions

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Chapter 6: Square Roots and Cube Roots > Finding Cube root through Prime Factorisation Method

Finding Cube root through Prime Factorisation Method

Consider 3375. Let's find its cube root using prime factorisation:

3375 = × × × × ×

= 33×53=3×53

Therefore, cube root of:

3375 = 33375 = × = (Enter factors in increasing order)

Similarly, to find 374088:

74088 = × × × × × × × × (Enter factors in increasing order)

= 23×33×73=2×3×73

Therefore,

374088 = × × =

Find the cube root of 8000

Prime factorisation of 8000 = × × × × × × × × (Enter factors in increasing order)

So,

38000 = 2 × 2 × 5 =

Find the cube root of 13824 by prime factorisation method.

13824 = × × × × × × × × × × × (Enter factors in increasing order)

= 23×23×23×33

Therefore,

313824= 2 × 2 × 2 × 3 =

Instructions

State true or false: for any integer m, m2 < m3. Why?

Let's consider four cases to check if the statement is true for all values. Case (i): m > 1 , Case (ii): m = 1 , Case (iii): m = 0 and Case (iv): m < 0.
Case (i) : m > 1. The statement is because m3 is than m2.
If m = 2, then 22 = 4 and 23 = 8 which gives us 4 8. Thus, m2 > m3.
Case (ii): m = 1. The statement is because m2 m3.
For m = 1, then 12 = 13.
Case (iii): m = 0. The statement is because m2 = m3.
For m = 0, then 02 = 03.
Case (iv): m < 0 (Negative Integers). The statement is because m3 m2.
If m = −2, then −22 = 4 and −23 = −8. So, 4 −8.
Thus, the statement is not true for all integers m.