Answer
$\dfrac{9}{8x^{5}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
(2x^3y^{-4})^{-3}(3x^2y^{-6})^2
,$ use the laws of exponents.
$\bf{\text{Solution Details:}}$
Using the extended Power Rule of the laws of exponents, which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(2x^3y^{-4})^{-3}(3x^2y^{-6})^2
\\\\
\left( 2^{-3}x^{3(-3)}y^{-4(-3)} \right) \left( 3^2x^{2(2)}y^{-6(2)} \right)
\\\\
\left( 2^{-3}x^{-9}y^{12} \right) \left( 3^2x^{4}y^{-12} \right)
.\end{array}
Using the Product Rule of the laws of exponents, which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( 2^{-3}x^{-9}y^{12} \right) \left( 3^2x^{4}y^{-12} \right)
\\\\=
(2^{-3})(3^2)x^{-9+4}y^{12+(-12)}
\\\\=
(2^{-3})(9)x^{-5}y^{0}
.\end{array}
Since any expression (except $0$) raised to the $0$ power is $1,$ then the expression above simplifies to
\begin{array}{l}\require{cancel}
(2^{-3})(9)x^{-5}y^{0}
\\\\
(2^{-3})(9)x^{-5}(1)
\\\\
(2^{-3})(9)x^{-5}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents, which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(2^{-3})(9)x^{-5}
\\\\=
\dfrac{9}{2^{3}x^{5}}
\\\\=
\dfrac{9}{8x^{5}}
.\end{array}
