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Powers and Roots Practice Test 2

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Question : 1 of 40

Marks: +1, -0

The value of

\frac{(0.96)^3 - (0.1)^3}{(0.96)^2 + (0.096) + (0.1)^2}

is:

0.86
1.06
0.95
0.97

Solution:

Concept:

Use the algebraic identity for the difference of cubes.

Explanation:

Let

a=0.96

and

b=0.1

.

The numerator is

a^3-b^3

.

Factorise it using

a^3-b^3=(a-b)(a^2+ab+b^2)

.

Here,

ab=0.96 \times 0.1=0.096

.

So the denominator is

a^2+ab+b^2

.

Cancel the common factor

a^2+ab+b^2

.

This leaves

a-b

.

Thus,

0.96-0.1=0.86

.

Answer:

0.86

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