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: 38

Answer

$-2x^{3}$

Work Step by Step

$\sqrt[5] (-32x^{15})=-2x^{3}$, because $(-2x^{3})^{5}=-2x^{3}\times -2x^{3}\times -2x^{3}\times -2x^{3}\times -2x^{3}=$ $(-2\times-2\times-2\times-2\times-2)\times x^{3+3+3+3+3}=-32x^{15}$
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.