## Precalculus (6th Edition) Blitzer

$\frac{x^2}{y^3}$
We follow the laws of exponents to simplify the expression: $$(x^{1/3}y^{-1/2})^6$$$$(x^{1/3})^6(y^{-1/2})^6$$$$x^2y^{-3}$$$$\frac{x^2}{y^3}$$