## College Algebra (11th Edition)

$\dfrac{m\sqrt[3]{n^2}}{n}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{\sqrt[3]{mn}\cdot\sqrt[3]{m^2}}{\sqrt[3]{n^2}} ,$ use the laws of radicals and the laws of exponents. Rationalize the denominator when necessary. $\bf{\text{Solution Details:}}$ 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{\sqrt[3]{mn(m^2)}}{\sqrt[3]{n^2}} .\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{mn(m^2)}{n^2}} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{m^{1+2}n}{n^2}} \\\\= \sqrt[3]{\dfrac{m^{3}\cancel{n}}{n^{\cancel{2}1}}} \\\\= \sqrt[3]{\dfrac{m^{3}}{n}} .\end{array} Rationalizing the denominator results to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{m^{3}}{n}\cdot\dfrac{n^2}{n^2}} \\\\= \sqrt[3]{\dfrac{m^{3}n^2}{n^3}} \\\\= \sqrt[3]{\dfrac{m^{3}}{n^3}\cdot n^2} \\\\= \sqrt[3]{\left(\dfrac{m}{n}\right)^3\cdot n^2} \\\\= \dfrac{m}{n}\sqrt[3]{n^2} \\\\= \dfrac{m\sqrt[3]{n^2}}{n} .\end{array}