Intermediate Algebra (6th Edition)

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

Chapter 8 - Cumulative Review - Page 534: 41c

Answer

$\frac{\sqrt[3] 4}{2}$

Work Step by Step

First, we simplify the given expression, $\sqrt[3] \frac{1}{2}=\frac{\sqrt[3] 1}{\sqrt[3] 2}=\frac{1}{\sqrt[3] 2}$ Next, we rationalize its denominator by multiplying the fraction by a term that will remove the rational exponent in the denominator, $\frac{1}{\sqrt[3] 2}\times\frac{\sqrt[3] 4}{\sqrt[3] 4}$ =$\frac{1\times\sqrt[3] 4}{\sqrt[3] 2\times\sqrt[3] 4}$ =$\frac{\sqrt[3] 4}{\sqrt[3] {2\times4}}$ =$\frac{\sqrt[3] 4}{\sqrt[3] {8}}$ =$\frac{\sqrt[3] 4}{2}$
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.