Answer
$\frac{16}{9}$
Work Step by Step
$(\frac{27}{64})^{-\frac{2}{3}}$
Using $a^{-\frac{m}{n}} = (\frac{1}{a^{\frac{m}{n}}})$
$(\frac{27}{64})^{-\frac{2}{3}} = (\frac{1}{\frac{27}{64}^{\frac{2}{3}}})$
$ = (\frac{1}{(\frac{3^{3}}{4^{3}})^{\frac{2}{3}}})$
$ = (\frac{1}{((\frac{3^{3}}{4^{3}})^{\frac{1}{3}})^{2}})$
$ = (\frac{1}{(\frac{3}{4})^{2}})$
$ = (\frac{1}{\frac{9}{16}})$
$ = \frac{16}{9}$
