## Intermediate Algebra for College Students (7th Edition)

$\displaystyle \frac{5\sqrt[4]{x^{2}y}}{xy^{2}}$
In the denominator, apply $\sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab},\qquad$ and $\sqrt[n]{a^{n}}=(\sqrt[n]{a})^{n}=a$ (for positive a). $\sqrt[4]{x^{2}y^{7}}=\sqrt[4]{y^{4}\times x^{2}y^{3}}=\sqrt[4]{y^{4}}\times\sqrt[4]{x^{2}y^{3}}=y\sqrt[4]{x^{2}y^{3}}$ we rationalize so that we get $\sqrt[4]{x^{4}y^{4}}$=$xy$ in the numerator $\displaystyle \frac{5}{y\sqrt[4]{x^{2}y^{3}}} \displaystyle \color{red}{ \cdot\frac{\sqrt[4]{x^{2}y}}{\sqrt[4]{x^{2}y}} }\qquad$ (rationalize) $=\displaystyle \frac{5\sqrt[4]{x^{2}y}}{y\sqrt[4]{x^{4}y^{4}}}$ $=\displaystyle \frac{5\sqrt[4]{x^{2}y}}{y\times xy}$ $=\displaystyle \frac{5\sqrt[4]{x^{2}y}}{xy^{2}}$