Intermediate Algebra (6th Edition)

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

Chapter 11 - Cumulative Review - Page 670: 14a



Work Step by Step

Rationalizing the denominator of $ \sqrt[3]{\dfrac{4}{3x}} $ results to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{4}{3x}\cdot\dfrac{9x^2}{9x^2}} \\\\= \sqrt[3]{\dfrac{36x^2}{27x^3}} \\\\= \dfrac{\sqrt[3]{36x^2}}{3x} .\end{array}
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