Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Review Exercises: 32

Answer

$\frac{1}{(\sqrt[3]3a+b)^{5}}$

Work Step by Step

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, $\frac{1}{(3a+b)^{\frac{5}{3}}}=\frac{1}{\sqrt[3](3a+b)^{5}}=\frac{1}{(\sqrt[3]3a+b)^{5}}$.
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