## Algebra 2 Common Core

$\dfrac{9y^6}{x^8}$
RECALL: (i) $(a^m)^n = a^{mn}$ (ii) $(ab)^m = a^mb^m$ (iii) $a^{-m} = \dfrac{1}{a^m}, a \ne0$ Use the rules above to have $3^2(x^{-4})^2(y^3)^2 \\=9x^{-4(2)}y^{3(2)} \\=9x^{-8}y^6 \\=\dfrac{9y^6}{x^8}.$