Beginning Algebra With Applications, Chapter 7, 7.4, Section 7.4, Problem 70

Simplify: $\displaystyle (3ab^{-2}) (2a^{-1}b)^{-3}$


$
\begin{equation}
\begin{aligned}

(3ab^{-2}) (2a^{-1}b)^{-3} =& (3ab^{-2}) \left( 2^{-3} a^{-1(-3)} b^{-3} \right)
&& \text{Use the rule for Simplifying Power of Products}
\\
\\
=& (3ab^{-2}) (2^{-3} a^3 b^{-3})
&& \text{Simplify}
\\
\\
=& 3a \left( \frac{1}{b^2} \right) \left( \frac{1}{2^3} \right) (a^3) \left( \frac{1}{b^3} \right)
&& \text{Write the expression with only positive exponents}
\\
\\
=& \frac{3a a^3}{2^3 b^2 b^3}
&& \text{Simplify}
\\
\\
=& \frac{3 a^{1+3}}{2^3 b^{2+3}}
&& \text{Multiply variables with same bases by adding their exponents}
\\
\\
=& \frac{3a^4}{8b^5}
&& \text{Simplify}

\end{aligned}
\end{equation}
$