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

$$=-\frac{x^{21}}{8y^{27}}$$
$$(\frac{-20x^{-2}y^3}{\:10x^5y^{-6}})^{-3}$$ Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m-n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m-n}}$ where $m>n$ Thus, the term in the parentheses can be simplified as: $$\frac{-20x^{-2}y^3}{\:10x^5y^{-6}}$$ $$\frac{-2y^{3-(-6)}}{x^{5-(-2)}}$$ $$\frac{-2y^{9}}{x^{7}}$$ The whole expression can now be written as: $$(\frac{-2y^{9}}{x^{7}})^{-3}$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$(\frac{-2y^{9}}{x^{7}})^{-3} = \frac{1}{(\frac{-2y^{9}}{x^{7}})^{3}}$$ $$=(\frac{x^{7}}{-2y^{9}})^{3}$$ Recall quotient-to-power rule: $(\frac{a}{b})^{n} = \frac{a^{n}}{b^{n}}$ Thus, $$(\frac{x^{7}}{-2y^{9}})^{3} = \frac{x^{7(3)}}{-2^{3}y^{9(3)}}$$ $$=\frac{x^{21}}{-8y^{27}}$$ $$=-\frac{x^{21}}{8y^{27}}$$