Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review: 35



Work Step by Step

$(\frac{8}{27})^{-\frac{2}{3}}=\frac{1}{(\frac{8}{27})^{\frac{2}{3}}}=\frac{1}{\sqrt[3] ((\frac{8}{27})^{2})}=\frac{1}{\sqrt[3] (\frac{64}{27\times27})}=\frac{1}{\frac{4}{3\times3}}=\frac{1}{\frac{4}{9}}=\frac{9}{4}$ We know that $\sqrt[3] 64=4$, because $4^{3}=64$. We know that $\sqrt[3] 27=3$, because $3^{3}=27$.
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