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

$\dfrac{\sqrt[3]{2xy^2}}{xy}$
Multiplying the given expression, $\sqrt[3]{\dfrac{2}{x^2y}} ,$ by an expression equal to $1$ such that the denominator becomes a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{2}{x^2y}} \\\\= \sqrt[3]{\dfrac{2}{x^2y}\cdot\dfrac{xy^2}{xy^2}} \\\\= \sqrt[3]{\dfrac{2xy^2}{x^3y^3}} .\end{array} Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{2xy^2}{x^3y^3}} \\\\= \dfrac{\sqrt[3]{2xy^2}}{\sqrt[3]{x^3y^3}} \\\\= \dfrac{\sqrt[3]{2xy^2}}{\sqrt[3]{(xy)^3}} \\\\= \dfrac{\sqrt[3]{2xy^2}}{xy} .\end{array}