Question

The value of $\frac{{\left(0.96\right)}^3-{\left(0.1\right)}^3}{{\left(0.90\right)}^2+\left(0.096\right)+0.01}$ is

• A. 0.86
• B. 1.06
• C. 0.95
• D. 0.97

Answer 1

Answer: (A)

0.86

$\Rightarrow \frac{{\left(0.96\right)}^3-{\left(0.1\right)}^3}{{\left(0.96\right)}^2+\left(0.096\right)+0.01}$

$\Rightarrow \frac{(0.96-0.1)[{\left(0.96\right)}^2+(0.96\times 0.1)+{\left(0.1\right)}^2]}{[{\left(0.96\right)}^2+\left(0.096\right)+\left(0.01\right)]}[\because a^3-b^3=(a-b)(a^2+ab+b^2)]$

$\Rightarrow \frac{0.86[{\left(0.96\right)}^2+\left(0.096\right)+\left(0.01\right)]}{[{\left(0.96\right)}^2+\left(0.096\right)+\left(0.01\right)]}$

$\Rightarrow 0.86$