Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.1 - Radicals and Radical Functions - Exercise Set - Page 417: 67



Work Step by Step

$\sqrt[5] (-32x^{10}y^{5})=-2x^{2}y$, because $(-2x^{2}y)^{5}=-2x^{2}y\times -2x^{2}y\times -2x^{2}y\times -2x^{2}y\times -2x^{2}y=(-2\times-2\times-2\times-2\times-2)\times x^{2+2+2+2+2}\times y^{1+1+1+1+1}=-32x^{10}y^{5}$
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