## Intermediate Algebra (12th Edition)

$\frac{1}{8}$
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 $2^{5}=32$