Answer
$-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{x^{2}y^{4}}$
Work Step by Step
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)
$=-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{xy^{3}\sqrt[3]{x^{3}y^{3}}}$
$=-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{xy^{3}\times xy}$
$=-\displaystyle \frac{5\sqrt[3]{x^{2}y}}{x^{2}y^{4}}$
