## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 10 - Exponents and Radicals - 10.4 Dividing Radical Expressions - 10.4 Exercise Set - Page 653: 49

#### Answer

$\dfrac{\sqrt[3]{75ac^2}}{5c}$

#### Work Step by Step

Rationalizing the denominator of the given expression, $\dfrac{\sqrt[3]{3a}}{\sqrt[3]{5c}} ,$ we find: \begin{array}{l}\require{cancel} \dfrac{\sqrt[3]{3a}}{\sqrt[3]{5c}}\cdot\dfrac{\sqrt[3]{25c^2}}{\sqrt[3]{25c^2}} \\\\= \dfrac{\sqrt[3]{75ac^2}}{\sqrt[3]{125c^3}} \\\\= \dfrac{\sqrt[3]{75ac^2}}{\sqrt[3]{(5c)^3}} \\\\= \dfrac{\sqrt[3]{75ac^2}}{5c} .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.