Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Review Exercises - Page 143: 58

Answer

$3\sqrt[3] 3$

Work Step by Step

$\sqrt[3] {81}=\sqrt[3] {27\times3}=\sqrt[3] {27}\times\sqrt[3] 3=3\sqrt[3] 3$
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