Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.3 - Simplifying Radical Expressions - Exercise Set: 34

Answer

$3\sqrt [3] 4$

Work Step by Step

$\sqrt [3] 108=\sqrt [3] 27\times\sqrt [3] 4=3\sqrt [3] 4$ We know that $\sqrt[3] 27=3$, because $3^{3}=27$.
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