College Algebra, Exercise P, Exercise P.4, Section Exercise P.4, Problem 56

Simplify the expression $\displaystyle \left( 3x^{-1} y^{-2} \right)^{-2} \left( x^2y^5 \right)^{-1}$ and eliminate any negative exponents.

$
\begin{equation}
\begin{aligned}
\left( 3x^{-1} y^{-2} \right)^{-2} \left( x^2y^5 \right)^{-1} &= \frac{1}{\left( x^2 y^5 \right)\left( 3x^{-1}y^{-2} \right) ^2} && \text{Definition of Negative Exponent } a^{-n} = \frac{1}{a^n}\\
\\
&= \frac{1}{\left( x^2y^5 \right)\left[ 3^2 (x^{-1})^2(y^{-2})^2\right]} && \text{Law: } (a^m)^n = a^{mn}\\
\\
&= \frac{1}{\left( x^2 y^5\right) \left( 9x^{-2} y^{-4} \right)} && \text{Definition of negative exponent } a^{-n} = \frac{1}{a^n}\\
\\
&= \frac{1}{\left( x^2 y^5 \right) \left( \frac{9}{x^2y^4} \right)} && \text{Law: } \frac{a^m}{a^n} = a^{m-n}\\
\\
&= \frac{1}{9x^{2-2}y^{5-4}} && \text{Simplify}\\
\\
&= \frac{1}{9x^0y^1} && \text{Simplify}\\
\\
&= \frac{1}{9y}
\end{aligned}
\end{equation}
$