College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Summary, Review, and Test - Review Exercises - Page 90: 59



Work Step by Step

$\sqrt[3] y^{5}=\sqrt[3](y^{3}\times y^{2})=\sqrt[3]y^{3}\times\sqrt[3]y^{2}=y\sqrt[3]y^{2}$ We know that $\sqrt[3] y^{3}=y$, because $(y)^{3}=y^{3}$.
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