The value of $\dfrac{(0.96)^3-(0.1)^3}{(0.96)^2+0.096-(0.1)^2}$ is

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0.86

0.95

0.97

1.06

Explanation:

Given expression

=$ \dfrac{(0.96)^3- (0.1)^3}{(0.96)^2+ (0.96 \times 0.1) + (0.1)^2} $

= $ \left(\dfrac{a^3- b^3}{a^2 + ab + b^2} \right) $

= $\left( a - b \right)$= (0.96 - 0.1)= 0.86

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The value of $\dfrac{0.1 \times 0.1 \times 0.1 + 0.02 \times 0.02 \times 0.02}{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}$ is

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