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