Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.3 - Simplifying Radical Expressions - Exercise Set - Page 433: 42



Work Step by Step

$\sqrt [3] (64y^{9})=\sqrt [3] 64 \times\sqrt[3] (y^{9}))=4y^{3}$ We know that $\sqrt [3] 64=4$, because $4^{3}=64$. We also know that $\sqrt[3] (y^{9})=y^{3}$, because $(y^{3})^{3}=y^{3\times3}=y^{9}$
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