#### Find the value of $\frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}$.

Solution. We have, $\frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}$
We know that
216 = 6.6.6 = 63
256 = 4.4.4.4 = 44
243 = 3.3.3.3.3 = 35
So, $\frac{4}{\left ( 216 \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 256 \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 243 \right )^{\frac{-1}{5}}}$

$= \frac{4}{\left ( 6^{3} \right )^{\frac{-2}{3}}}+\frac{1}{\left ( 4^{4} \right )^{\frac{-3}{4}}}+\frac{2}{\left ( 3 ^{5}\right )^{-\frac{1}{5}}}$
$= \frac{4}{\left ( 6 \right )^{3\times \frac{-2}{3}}}+\frac{1}{\left ( 4 \right )^{4\times \frac{-3}{4}}}+\frac{2}{\left ( 3 \right )^{5\times \frac{-1}{5}}}$
$\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= \frac{4}{6^{-2}}+\frac{1}{4^{-3}}+\frac{2}{3^{-1}}$
= 4 × 62 + 43 + 2 × 3
$\because \frac{1}{\left ( a \right )^{-n}}= \left ( a \right )^{n}$
= 4 × 36 + 64 + 6
= 144 + 70
= 214