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

The answer is $\frac{5\sqrt[3] {4xy^2}}{2xy}$.
The given expression is $\frac{5}{\sqrt[3] {2x^2y}}$ To rationalize the denominator we multiply and divide by $\sqrt[3] {2^2xy^2}$ $\frac{5}{\sqrt[3] {2x^2y}} \cdot \frac{\sqrt[3] {2^2xy^2}}{\sqrt[3] {2^2xy^2}}$ Simplify. $\frac{5\sqrt[3] {2^2xy^2}}{2xy}$. $\frac{5\sqrt[3] {4xy^2}}{2xy}$.