## Intermediate Algebra (12th Edition)

$5\sqrt[3]{3}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{15}{\sqrt[3]{9}} ,$ multiply by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ 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{15}{\sqrt[3]{3^2}} \\\\= \dfrac{15}{\sqrt[3]{3^2}}\cdot\dfrac{\sqrt[3]{3}}{\sqrt[3]{3}} \\\\= \dfrac{15\sqrt[3]{3}}{\sqrt[3]{3^2}(\sqrt[3]{3})} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{15\sqrt[3]{3}}{\sqrt[3]{3^2(3)}} \\\\= \dfrac{15\sqrt[3]{3}}{\sqrt[3]{3^3}} \\\\= \dfrac{15\sqrt[3]{3}}{3} \\\\= \dfrac{\cancel{3}(5)\sqrt[3]{3}}{\cancel{3}} \\\\= 5\sqrt[3]{3} .\end{array}