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

If $(x + \frac{1}{x})^2 = 3$,then $(x^3 + \frac{1}{x^3})$ is equal to

If $2x + \frac{2}{x} = 1$,then the value of $x^{3} + \frac{1}{x^{3}}$ is

$7 \frac{1}{2}-\left[2 \frac{1}{4} \div\left\{1 \frac{1}{4}-x\left(1 \frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right\}\right]=3$. Find the value of $x$.

If $\frac{x^{24}+1}{x^{12}}=7$,then find the value of $\frac{x^{72}+1}{x^{36}}$.

Simplify:
$\frac{3}{10} \div \frac{3}{7} \text{ of } \left(2 \frac{3}{10} + 2 \frac{3}{5}\right) + \frac{1}{5} \div 1 \frac{2}{5} - \frac{2}{7}$