Answer
-2y
Work Step by Step
$\sqrt[5] (-32y^{5})=\sqrt[5] (-32)\times\sqrt[5] (y^{5})=-2y$
We know that $\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$. Also, we know that $\sqrt[5] (y^{5})=y$, because $(y)^{5}=y^{5}$
Intermediate Algebra (6th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321785045
ISBN 13:
978-0-32178-504-6
Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents - Page 446: 7
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