## Intermediate Algebra (12th Edition)

$-\dfrac{\sqrt[3]{6xy}}{y}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $-\sqrt[3]{\dfrac{6x}{y^2}} ,$ 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{6x}{y^2}\cdot\dfrac{y}{y}} \\\\= -\sqrt[3]{\dfrac{6xy}{y^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} -\dfrac{\sqrt[3]{6xy}}{\sqrt[3]{y^3}} \\\\= -\dfrac{\sqrt[3]{6xy}}{y} .\end{array}