ICSE Class 8 Maths Number Systems Chapter 10 Powers and Roots PDF Download

Powers and Roots

Cubes and Cube Roots - Table of Squares and Cubes

Introduction

In the previous chapter we have learnt how to find square roots of numbers by prime factorisation as well as division method.

We have learnt how to find cube roots of perfect cubes by prime factorisation method in our previous class.

Cubes and Cube Roots

When a number is multiplied by its square, it is said to be cubed.

Example 1: \(5 \times 5^2 = 5 \times 5 \times 5 = 5^3 = 125\)

Example 2: \(18 \times 18^2 = 18 \times 18 \times 18 = 18^3 = 5832\)

When a number 'a' multiplied by its square gives a certain product \(a^3\), the number a is known as the cube root of the product \(a^3\).

\(\sqrt[3]{a^3} = \sqrt[3]{a \times a \times a} = a\) where the sign '\(\sqrt[3]{}\)' denotes cube root.

Example 3: Find the cube root of 27000.

Prime factorisation of 27000

\(= 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5\)

Triplets of three identical factors

\(= \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}\)

Product of one factor from each triplet gives the cube root.

\(\sqrt[3]{27000} = \sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5}\)

\(= 2 \times 3 \times 5 = 30\)

Thus, the cube root of 27000 is 30.

Example 4: Find the cube root of 250047.

Prime factorisation of 250047

\(= 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7\)

Triplets of three identical factors

\(= 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7\)

Product of one factor from each triplet gives the cube root.

\(\sqrt[3]{250047} = \sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7}\)

\(= 3 \times 3 \times 7 = 63\)

Thus, the cube root of 250047 is 63.

Try this!

Find the cube root of 117649.

Table of Squares and Cubes

In this chapter, a table showing the squares, cubes, square roots, and cube roots of natural numbers up to 50 is given. A table like this is an 'aid' to make calculations easier just the way an electronic calculator is. It is a 'ready reckoner' that can be consulted whenever calculations involve indices 2, 3, \(\frac{1}{2}\), or \(\frac{1}{3}\).

Teacher's Note

Tables of squares and cubes serve as quick reference tools, much like looking up contact information in a phone book or consulting a conversion chart while cooking.

Using The Table

Example 5: Evaluate \(\sqrt{38} - \sqrt[3]{38}\).

Run your finger down the column for natural numbers 'n' and stop at 38. Move right along the row of 38 and stop at the column for the square root of n, correct up to 3 decimal places. Read the value of \(\sqrt{38}\). Similarly, read the value of \(\sqrt[3]{38}\) under the column for \(\sqrt[3]{n}\).

\(\sqrt{38} - \sqrt[3]{38} = 6.164 - 3.362 = 2.802\)

Example 6: Evaluate \(2\sqrt{19} + 3\sqrt[3]{15}\).

From the table,

\(\sqrt{19} = 4.359\) and \(\sqrt[3]{15} = 2.466\)

Then \(2 \times 4.359 + 3 \times 2.466 = 8.718 + 7.398 = 16.116\)

Example 7: Use the table to verify

\(45^3 - 20^3 = (45 - 20)(45^2 + 45 \times 20 + 20^2)\)

Reading the values of the squares and cubes from the given table,

\(91125 - 8000 = 25(2025 + 900 + 400)\)

\(\Rightarrow 83125 = 25 \times 3325\)

\(\Rightarrow 83125 = 83125\)

Thus the given statement is verified.

Example 8: Evaluate \(\sqrt[3]{930}\), correct up to 3 decimal places.

\(\sqrt[3]{930} = \sqrt[3]{31 \times 30}\)

\(= \sqrt[3]{31} \times \sqrt[3]{30}\)

\(= 3.141 \times 3.107\)

\(= 9.759087\)

\(= 9.759\) correct up to 3 decimal places

Teacher's Note

Using reference tables to solve problems is like using GPS to navigate - it saves time and reduces errors when calculations are frequent.

Exercise 10.1

1. Read the values of the following from the table.

(i) \(23^2\) (ii) \(49^2\) (iii) \(28^2\) (iv) \(39^2\) (v) \(47^2\) (vi) \(19^3\) (vii) \(27^3\) (viii) \(36^3\) (ix) \(41^3\) (x) \(48^3\) (xi) \(\sqrt{15}\) (xii) \(\sqrt{26}\) (xiii) \(\sqrt{29}\) (xiv) \(\sqrt{37}\) (xv) \(\sqrt{47}\) (xvi) \(\sqrt[3]{13}\) (xvii) \(\sqrt[3]{31}\) (xviii) \(\sqrt[3]{22}\) (xix) \(\sqrt[3]{50}\) (xx) \(\sqrt[3]{34}\)

2. Use the table to verify

\(35^3 + 15^3 = (35 + 15)(35^2 - 35 \times 15 + 15^2)\)

3. Evaluate the following using the table.

(i) \(19^2 + 10^3\) (ii) \(45^3 - 30^3\) (iii) \(12^3 + 9^3 - 11^3\) (iv) \(14^3 + 15^3 - 16^3\) (v) \(\sqrt{12} + \sqrt{15}\) (vi) \(\sqrt{27} - \sqrt{7}\) (vii) \(\sqrt[3]{43} + \sqrt[3]{34}\) (viii) \(\sqrt[3]{37} - \sqrt[3]{3}\) (ix) \(\sqrt{2} + \sqrt{2} - \sqrt{3}\) (x) \(\sqrt[3]{49} + \sqrt[3]{50} - \sqrt{50}\) (xi) \(\sqrt{57}\) (xii) \(\sqrt[3]{430}\) (xiii) \(\sqrt[3]{99}\) (xiv) \(\sqrt[3]{196}\)

Revision Exercise

1. Read the values of the following from the table.

(i) \(26^2\) (ii) \(43^3\) (iii) \(32^2\) (iv) \(23^3\) (v) \(\sqrt{41}\) (vi) \(\sqrt{33}\) (vii) \(\sqrt[3]{38}\) (viii) \(\sqrt[3]{46}\)

2. Use the table to verify.

(a) \(28^2 - 21^2 = (28 + 21)(28 - 21)\) (b) \(48^3 - 23^3 = (48 - 23)(48^2 + 48 \times 23 + 23^2)\)

3. Evaluate the following using the table.

(i) \(21^2 + 19^2\) (ii) \(49^2 - 29^2\) (iii) \(16^3 - 12^3\) (iv) \(18^3 + 19^3 - 20^3\) (v) \(\sqrt{33} + \sqrt{35}\) (vi) \(\sqrt{45} - 3\sqrt{5}\) (vii) \(\sqrt{16} - \sqrt{14}\) (viii) \(\sqrt[3]{50} + \sqrt[3]{49} - \sqrt[3]{30}\)

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