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: 49


$-2x^{2}\sqrt[5] y$

Work Step by Step

$\sqrt [5] (-32x^{10}y)=\sqrt [5] (-32\times x^{10}\times y)=\sqrt [5] (-32)\times \sqrt[5] (x^{10})\times \sqrt[5] y=-2x^{2}\sqrt[5] y$ We know that $\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$ and also that $\sqrt[5] (x^{10})=x^{2}$, because $(x^{2})^{5}=x^{2\times5}=x^{10}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.