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

$\dfrac{\sqrt[3]{2xy^2}}{2y}$
Multiplying by an expression equal to $1$ which will make the denominator a perfect power of the index, the given expression is equivalent to\begin{array}{l}\require{cancel} \dfrac{\sqrt[3]{x}}{\sqrt[3]{4y}} \\\\= \dfrac{\sqrt[3]{x}}{\sqrt[3]{4y}}\cdot\dfrac{\sqrt[3]{2y^2}}{\sqrt[3]{2y^2}} \\\\= \dfrac{\sqrt[3]{x(2y^2)}}{\sqrt[3]{4y(2y^2)}} \\\\= \dfrac{\sqrt[3]{2xy^2}}{\sqrt[3]{8y^3}} \\\\= \dfrac{\sqrt[3]{2xy^2}}{\sqrt[3]{(2y)^3}} \\\\= \dfrac{\sqrt[3]{2xy^2}}{2y} .\end{array}