## Intermediate Algebra (12th Edition)

$\dfrac{x^2\sqrt[3]{y^2}}{y}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\sqrt[3]{\dfrac{x^6}{y}} ,$ multiply the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{x^6}{y}\cdot\dfrac{y^2}{y^2}} \\\\= \sqrt[3]{\dfrac{x^6y^2}{y^3}} .\end{array} Rewriting the radicand using an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{x^6}{y^3}\cdot y^2} \\\\= \sqrt[3]{\left( \dfrac{x^2}{y} \right)^3\cdot y^2} .\end{array} Extracting the root of the radicand that is a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{x^2}{y}\sqrt[3]{y^2} \\\\= \dfrac{x^2\sqrt[3]{y^2}}{y} .\end{array}