Question

Question) Simplify :
$\left(i\right)\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}$
$$\begin{gathered} ii)\left(3^2-2^2\right)\times {\left(\frac{2}{3}\right)}^{-3} \\ (iii){\left\{{\left(\frac{1}{2}\right)}^{-1}\times {\left(-4\right)}^{-1}\right\}}^{-1} \\ (iv){\left[{\left\{{\left(\frac{-1}{4}\right)}^2\right\}}^{-2}\right]}^{-1} \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} 2)\left(3^2-2^2\right)\times {\left(\frac{2}{3}\right)}^{-3} \\ =\left(9-4\right)\times {\left(\frac{3}{2}\right)}^3 \\ =5\times \frac{9}{8} \\ =\frac{45}{8} \\ 4){\left[{\left\{{\left(\frac{-1}{4}\right)}^2\right\}}^{-2}\right]}^{-1} \\ ={\left(\frac{-1}{4}\right)}^4\left[\mathrm{as}{\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right] \\ =\frac{1}{256} \end{gathered}$$
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