Answer
$-\displaystyle \frac{6\sqrt[3]{xy}}{x^{2}y^{3}}$
Work Step by Step
Simplify the denominator using the following:
$\sqrt[3]{-8}=\sqrt[3]{(-2)^{3}}=-2$
$\sqrt[3]{x^{5}}=\sqrt[3]{x^{3}\times x^{2}}=\sqrt[3]{x^{3}}\times\sqrt[3]{x^{2}}=x\sqrt[3]{x^{2}}$
$\sqrt[3]{y^{8}}=\sqrt[3]{(y^{2})^{3}\times y^{2}}=\sqrt[3]{(y^{2})^{3}}\times\sqrt[3]{y^{2}}=y^{2}\sqrt[3]{y^{2}}$
$\displaystyle \frac{12}{\sqrt[3]{-8x^{5}y^{8}}}=\frac{12}{-2\times x\sqrt[3]{x^{2}}\times y^{2}\sqrt[3]{y^{2}}}=-\frac{12}{2xy^{2}\sqrt[3]{x^{2}y^{2}}}$
...(cancel 2)
We rationalize with the goal of obtaining $\sqrt[3]{x^{3}y^{3}}$
$=-\displaystyle \frac{6}{xy^{2}\sqrt[3]{x^{2}y^{2}}} \displaystyle \color{red}{ \cdot\frac{\sqrt[3]{xy}}{\sqrt[3]{xy}} }\qquad$ (rationalize)
$=-\displaystyle \frac{6\sqrt[3]{xy}}{xy^{2}\sqrt[3]{x^{3}y^{3}}}$
$=-\displaystyle \frac{6\sqrt[3]{xy}}{xy^{2}\times xy}$
$=-\displaystyle \frac{6\sqrt[3]{xy}}{x^{2}y^{3}}$
