## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 7 - Review: 105

#### Answer

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

#### Work Step by Step

Multiplying both the numerator and the denominator by a factor that will make the denominator a perfect power of the radical, the rationalized-denominator form of the given expression, $\dfrac{5}{\sqrt[3]{4}} ,$ is \begin{array}{l}\require{cancel} \dfrac{5}{\sqrt[3]{4}}\cdot\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}} \\\\= \dfrac{5\sqrt[3]{2}}{\sqrt[3]{4(2)}} \\\\= \dfrac{5\sqrt[3]{2}}{\sqrt[3]{(2)^3}} \\\\= \dfrac{5\sqrt[3]{2}}{3} .\end{array}

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