Question

${\left[\frac{(169{)}^{-3}}{(196{)}^{-8}}\right]}^{\frac{1}{48}}$ is equal to _____.

Answer 1

${\left[\frac{(169{)}^{-3}}{(196{)}^{-8}}\right]}^{\frac{1}{48}}$
$={\left[\frac{({13}^2{)}^{-3}}{({14}^2{)}^{-8}}\right]}^{\frac{1}{48}}$
$={\left[\frac{{13}^{-6}}{{14}^{-16}}\right]}^{\frac{1}{48}}$

$=\frac{{13}^{-6\times \frac{1}{48}}}{{14}^{-16\times \frac{1}{48}}}$

$=\frac{{13}^{-\frac{1}{8}}}{{14}^{-\frac{1}{3}}}$

$=\frac{{14}^{\frac{1}{3}}}{{13}^{\frac{1}{8}}}$
Hence, the required result is $\frac{{14}^{\frac{1}{3}}}{{13}^{\frac{1}{8}}}$ .