Answer
$-\dfrac{27a^{6}}{b^{18}}$
Work Step by Step
Using the laws of exponents, the given expression, $
\left( \dfrac{-4a^3b^2}{12a^5b^{-4}} \right)^{-3}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(-4)^{-3}a^{3(-3)}b^{2(-3)}}{12^{-3}a^{5(-3)}b^{-4(-3)}}
\\\\=
\dfrac{(-4)^{-3}a^{-9}b^{-6}}{12^{-3}a^{-15}b^{12}}
\\\\=
\dfrac{(-4)^{-3}a^{-9-(-15)}b^{-6-12}}{(-4)^{-3}(-3)^{-3}}
\\\\=
\dfrac{\cancel{(-4)^{-3}}a^{-9+15}b^{-6-12}}{\cancel{(-4)^{-3}}(-3)^{-3}}
\\\\=
\dfrac{a^{6}b^{-18}}{(-3)^{-3}}
\\\\=
\dfrac{(-3)^{3}a^{6}}{b^{18}}
\\\\=
\dfrac{-27a^{6}}{b^{18}}
\\\\=
-\dfrac{27a^{6}}{b^{18}}
.\end{array}
