Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents - Page 447: 36

Answer

$\dfrac{5\sqrt[3]{4x}}{2x}$

Work Step by Step

Multiplying both the numerator and the denominator by $ \sqrt[3]{4x} $, then the rationalized-denominator form of the given expression, $ \dfrac{5}{\sqrt[3]{2x^2}} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{5}{\sqrt[3]{2x^2}}\cdot\dfrac{\sqrt[3]{4x}}{\sqrt[3]{4x}} \\\\= \dfrac{5\sqrt[3]{4x}}{\sqrt[3]{8x^3}} \\\\= \dfrac{5\sqrt[3]{4x}}{\sqrt[3]{(2x)^3}} \\\\= \dfrac{5\sqrt[3]{4x}}{2x} .\end{array}
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