Answer
$\frac{1}{8}$
Work Step by Step
We know that $a^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}$, where $a^{\frac{m}{n}}$ is a real number.
Therefore, $32^{-\frac{3}{5}}=\frac{1}{32^{\frac{3}{5}}}$
We know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=\sqrt[n] (a)^{m}$, where all indicated roots are real numbers.
Therefore, $\frac{1}{32^{\frac{3}{5}}}=\frac{1}{\sqrt[5] 32^{3}}=\frac{1}{(\sqrt[5] 32)^{3}}=\frac{1}{2^{3}}=\frac{1}{8}$
$\sqrt[5] 32=2$, because $2^{5}=32$
