College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises - Page 76: 104



Work Step by Step

Getting the factor that is a perfect root of the index, the given expression, $ \sqrt[3]{16} ,$ simplifies to \begin{array}{l}\require{cancel} \sqrt[3]{8\cdot2} \\\\= \sqrt[3]{(2)^3\cdot2} \\\\= 2\sqrt[3]{2} .\end{array}
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