## Intermediate Algebra (12th Edition)

$\frac{1}{(\sqrt[3]3a+b)^{5}}$
We are given that $a^{-r}=\frac{1}{a^{r}}$, if $a$ is a real number and $r$ is a rational number. Therefore, $(3a+b)^{-\frac{5}{3}}=\frac{1}{(3a+b)^{\frac{5}{3}}}$. We are given that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$, if all indicated roots are real numbers. Therefore, $\frac{1}{(3a+b)^{\frac{5}{3}}}=\frac{1}{\sqrt[3](3a+b)^{5}}=\frac{1}{(\sqrt[3]3a+b)^{5}}$.