Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 24

Answer

$-2b^{2}\sqrt[5] 2$

Work Step by Step

$\sqrt[5] (-64b^{10})=\sqrt[5] (-32\times b^{10}\times2)=\sqrt[5] (-32)\times \sqrt[5] (b^{10})\times\sqrt[5] 2=-2b^{2}\sqrt[5] 2$ We know that $\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$. We know that $\sqrt[5] (b^{10})=b^{2}$, because $(b^{2})^{5}=b^{2\times5}=b^{10}$.
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