Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises - Page 459: 55

Answer

$4\sqrt[3] 2$

Work Step by Step

$\sqrt[3] 128=\sqrt[3] (64\times2)=\sqrt[3] 64\times\sqrt[3] 2=4\sqrt[3] 2$ We know that $\sqrt[3] 64=4$, because $4^{3}=64$.
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