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: 21

Answer

$\frac{1}{2}$

Work Step by Step

$\sqrt[3] \frac{1}{8}=\frac{1}{2}$, because $(\frac{1}{2})^{3}=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}$
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