College Algebra, Exercise P, Exercise P.4, Section Exercise P.4, Problem 54

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

$
\begin{equation}
\begin{aligned}
\left(2u^2 v^3 \right)^3 \left( 3u^{-3}v \right)^2 &= \left[ 2^3(u^2)^3(v^3)^3 \right] \left[ 3^2(u^{-3})^2 v^2 \right] && \text{Law: } (ab)^n = a^n b^n\\
\\
&= \left( 8u^6 v^9 \right) \left( 9u^{-6} v^2 \right) && \text{Law: } (a^m)^n = a^{mn}\\
\\
&= (8) (9) u^6 u^{-6} v^9 v^2 && \text{Group factors with same base}\\
\\
&= 72 u^{6+(-6)} v^{9+2} && \text{Law: } a^m a^n = a^{m+n}\\
\\
&= 72u^0 v^{11} && \text{Simplify}\\
\\
&= 73v^{11}
\end{aligned}
\end{equation}
$