Intermediate Algebra (6th Edition)

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

Chapter 7 - Review: 12



Work Step by Step

$\sqrt[5] (-32x^{15}y^{20})=\sqrt[5] (-32\times x^{15}\times y^{20})=\sqrt[5] (-32)\times\sqrt[5] (x^{15})\times\sqrt[5] (y^{20})=-2x^{3}y^{4}$ $\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$ $\sqrt[5] (x^{15})=x^{3}$, because $(x^{3})^{5}=x^{3\times5}=x^{15}$ $\sqrt[5] (y^{20})=y^{4}$, because $(y^{4})^{5}=y^{4\times5}=y^{20}$
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