Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.1 - Radicals and Radical Functions - Exercise Set: 74



Work Step by Step

$\sqrt[3] (\frac{64a^{3}}{b^{9}})=\frac{4a}{b^{3}}$, because $(\frac{4a}{b^{3}})^{3}=\frac{4a}{b^{3}}\times\frac{4a}{b^{3}}\times \frac{4a}{b^{3}}=\frac{(4\times4\times4)\times a^{1+1+1}}{b^{3+3+3}}=\frac{64a^{3}}{b^{9}}$ Therefore, $-\sqrt[3] (\frac{64a^{3}}{b^{9}})=-\frac{4a}{b^{3}}$. We are just taking the opposite of $\sqrt[3] (\frac{64a^{3}}{b^{9}})$.
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