## Intermediate Algebra (12th Edition)

$\dfrac{m^3\sqrt[3]{q^2}}{q}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\sqrt[3]{\dfrac{m^9}{q}} ,$ 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{m^9}{q}\cdot\dfrac{q^2}{q^2}} \\\\= \sqrt[3]{\dfrac{m^9q^2}{q^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{m^9}{q^3}\cdot q^2} \\\\= \sqrt[3]{\left( \dfrac{m^3}{q} \right)^3\cdot q^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{m^3}{q}\sqrt[3]{q^2} \\\\= \dfrac{m^3\sqrt[3]{q^2}}{q} .\end{array}