Intermediate Algebra (6th Edition)

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

Chapter 7 - Review: 11



Work Step by Step

$\sqrt[5] (32a^{5}b^{10})=\sqrt[5] (32\times a^{5}\times b^{10})=\sqrt[5] 32\times\sqrt[5] (a^{5})\times\sqrt[5] (b^{10})=2ab^{2}$ $\sqrt[5] 32=2$, because $2^{5}=32$ $\sqrt[5] (a^{5})=a$, because $(a)^{5}=a^{5}$ $\sqrt[5] (b^{10})=b^{2}$, because $(b^{2})^{5}=b^{2\times5}=b^{10}$
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