College Algebra, Exercise P, Exercise P.4, Section Exercise P.4, Problem 68

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

$
\begin{equation}
\begin{aligned}
\left( \frac{x^2y}{2y^3} \right)^{-2} &= \left( \frac{2y^3}{x^2y} \right)^2 && \text{Law: } \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n\\
\\
&= \frac{2^2(y^3)^2}{(x^2)^2(y)^2} && \text{Law: } (ab)^n = a^n b^n\\
\\
&= \frac{4y^6}{x^4y^2} && \text{Law: } (a^m)^n = a^{mn}\\
\\
&= \frac{4y^{6-2}}{x^4} && \text{Law: } \frac{a^m }{a^n} = a^{m-n}\\
\\
&= \frac{4y^4}{x^4}
\end{aligned}
\end{equation}
$