Intermediate Algebra (6th Edition)

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

Chapter 7 - Review - Page 469: 65

Answer

$3\sqrt[3] 6$

Work Step by Step

$\sqrt[3] 162=\sqrt[3] (27\times6)=\sqrt[3] 27\times\sqrt[3] 6=3\sqrt[3] 6$ We know that $\sqrt[3] 27=3$, because $3^{3}=27$.
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