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: 41


$2y^{2}\sqrt[3] (2y)$

Work Step by Step

$\sqrt [3](16y^{7})=\sqrt[3] (8y^{6}\times 2y)=\sqrt[3] (8y^{6})\times\sqrt[3] (2y)=2y^{2}\sqrt[3] (2y)$ We know that $\sqrt[3] (8y^{6})=2y^{2}$, because $(2y^{2})^{3}=(2\times2\times2)\times y^{2+2+2}=8y^{6}$.
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