Question

Simplify the following and express with positive index.

If ${\left(\frac{{27}^{-3}}{9^{-3}}\right)}^{\frac{1}{5}}$ is $\frac{1}{b^{\frac{3}{5}}}$, then $b$ is:

Answer 1

$(\frac{{27}^{-3}}{9^{-3}}{)}^{\frac{1}{5}}$
$=(\frac{27}{9}{)}^{-\frac{3}{5}}$
$=3^{-\frac{3}{5}}$
$=\frac{1}{3^{\frac{3}{5}}}$

Thus, $b=3$.