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

$x^{12}y^6$
To solve $(\frac{x^4}{y^{-2}})^3$, RECALL: (i) The quotients-to-powers rule states that: $(\frac{a}{b})^n=\frac{a^n}{b^n}$ (ii) The power-rule states that $(a^m)^n=a^{mn}$ (iii) The negative-exponent rule states that: $a^{−m} =\frac{1}{a^m}$ and $\frac{1}{a^{-m}} = a^{m}$ Hence, using quotients-to-powers rule and power rule: $(\frac{x^4}{y^{-2}})^3$ = $\frac{(x^{4})^3}{(y^{-2})^{3}}$ Using the power rule: $=\frac{x^{4\times3}}{y^{-2\times3}}$ $=\frac{x^{12}}{y^{-6}}$ Using negative-exponent rule, $=\frac{x^{12}}{y^{-6}} = x^{12}y^6$