College Algebra, Exercise P, Exercise P.4, Section Exercise P.4, Problem 70

Simplify the expression $\displaystyle \left( \frac{2a^{-1}b}{a^2b^{-3}} \right)^{-3}$ and eliminate any negative exponents.

$
\begin{equation}
\begin{aligned}
\left( \frac{2a^{-1}b}{a^2b^{-3}} \right)^{-3} &= \left( \frac{a^2b^{-3}}{2a^{-1}b} \right) && \text{Law: } \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n\\
\\
&= \frac{(a^2)^3(b^{-3})^3}{2^3 (a^{-1})^3(b)^3} && \text{Law: } (ab)^n = a^n b^n\\
\\
&= \frac{a^6b^{-9}}{8a^{-3}b^3} && \text{Law: } (a^m)^n = a^{mn}\\
\\
&= \frac{a^6 a^3}{8b^9 b^3} && \text{Law: } \frac{a^{-n}}{a^{-m}} = \frac{b^m}{a^n}\\
\\
&= \frac{a^{6+3}}{8b^{9+3}} && \text{Law: } a^m a^n = a^{m+n}\\
\\
&= \frac{a^9}{8b^{12}}
\end{aligned}
\end{equation}
$