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

$\dfrac{\sqrt[5]{9y^4}}{2xy}$
Rationalizing the denominator of the given expression, $\sqrt[5]{\dfrac{9}{32x^5y}} ,$ we find: \begin{array}{l}\require{cancel} \sqrt[5]{\dfrac{9}{32x^5y}\cdot\dfrac{y^4}{y^4}} \\\\= \sqrt[5]{\dfrac{9y^4}{32x^5y^5}} \\\\= \dfrac{\sqrt[5]{9y^4}}{\sqrt[5]{32x^5y^5}} \\\\= \dfrac{\sqrt[5]{9y^4}}{\sqrt[5]{(2xy)^5}} \\\\= \dfrac{\sqrt[5]{9y^4}}{2xy} .\end{array} * Note that it is assumed that all variables represent positive numbers.