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

$-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{x^{2}y^{4}}$
Simplify the denominator using the following: $\sqrt[3]{-27}=\sqrt[3]{(-3)^{3}}=-3$ $\sqrt[3]{x^{4}}=\sqrt[3]{x^{3}\times x}=\sqrt[3]{x^{3}}\times\sqrt[3]{x}=x\sqrt[3]{x}$ $\sqrt[3]{y^{11}}=\sqrt[3]{(y^{3})^{3}\times y^{2}}=\sqrt[3]{(y^{3})^{3}}\times\sqrt[3]{y^{2}}=y^{3}\sqrt[3]{y^{2}}$ $\displaystyle \frac{15}{\sqrt[3]{-27x^{4}y^{11}}}=\frac{15}{-3\times x\sqrt[3]{x}\times y^{3}\sqrt[3]{y^{2}}}=\quad$...(cancel $3$) $=-\displaystyle \frac{5}{xy^{3}\sqrt[3]{xy^{2}}}$ ...(cancel 2) We rationalize with the goal of obtaining $\sqrt[3]{x^{3}y^{3}}$ $=-\displaystyle \frac{5}{xy^{3}\sqrt[3]{xy^{2}}}$ $\displaystyle \color{red}{ \cdot\frac{\sqrt[3]{x^{2}y}}{\sqrt[3]{x^{2}y}} }\qquad$ (rationalize) $=-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{xy^{3}\sqrt[3]{x^{3}y^{3}}}$ $=-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{xy^{3}\times xy}$ $=-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{x^{2}y^{4}}$