## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\displaystyle \frac{x^{12}}{9y^{8}}$
... apply: $(abc)^{n}=a^{n}b^{n}c^{n}$ $=(-1)^{-2}(3)^{-2}(x^{-6})^{-2}(y^{4})^{-2}\qquad$ ... apply $(a^{m})^{n}=a^{mn}$, and $a^{-n}=\displaystyle \frac{1}{a^{n}}$ $=[(-1)^{2}]^{-1}(\displaystyle \frac{1}{3^{2}})x^{12}y^{-8}$ $=\displaystyle \frac{1^{-1}x^{12}}{3^{2}y^{8}}$ $=\displaystyle \frac{x^{12}}{9y^{8}}$