## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\sqrt[3]{x^{2}y^{}}$
Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the given expression is equivalent to \begin{array}{l}\require{cancel} \left( \sqrt[6]{x^2y}\right)^2 \\\\= (x^2y)^{2/6} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x^2y)^{2/6} \\\\= x^{2\cdot\frac{2}{6}}y^{\frac{2}{6}} \\\\= x^{\frac{4}{6}}y^{\frac{2}{6}} \\\\= x^{\frac{2}{3}}y^{\frac{1}{3}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the given expression is equivalent to \begin{array}{l}\require{cancel} x^{\frac{2}{3}}y^{\frac{1}{3}} \\\\= \sqrt[3]{x^{2}y^{}} .\end{array}