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

$\dfrac{\sqrt[3]{63xy^2}}{3y}$
Rationalizing the denominator of the given expression, $\dfrac{\sqrt[3]{7x}}{\sqrt[3]{3y}} ,$ we find: \begin{array}{l}\require{cancel} \dfrac{\sqrt[3]{7x}}{\sqrt[3]{3y}}\cdot\dfrac{\sqrt[3]{9y^2}}{\sqrt[3]{9y^2}} \\\\= \dfrac{\sqrt[3]{63xy^2}}{\sqrt[3]{27y^3}} \\\\= \dfrac{\sqrt[3]{63xy^2}}{\sqrt[3]{(3y)^3}} \\\\= \dfrac{\sqrt[3]{63xy^2}}{3y} .\end{array} * Note that it is assumed that all variables represent positive numbers.