Answer
$\dfrac{4}{xy^2}$
Work Step by Step
Using the laws of exponents, the given expression, $
\dfrac{(4x^{-2})^2(2y^3)}{8x^{-3}y^5}
$, is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{(4^{1(2)}x^{-2(2)})(2y^3)}{8x^{-3}y^5}
&(\text{use }(x^my^n)^p=x^{mp}y^{np})
\\\\&=
\dfrac{(4^{2}x^{-4})(2y^3)}{8x^{-3}y^5}
\\\\&=
\dfrac{(16x^{-4})(2y^3)}{8x^{-3}y^5}
\\\\&=
\dfrac{32x^{-4}y^3}{8x^{-3}y^5}
\\\\&=
(32\div8)(x^{-4-(-3)}y^{3-5})
&(\text{use }\dfrac{x^m}{x^n}=x^{m-n})
\\&=
4(x^{-4+3}y^{3-5})
\\&=
4(x^{-1}y^{-2})
\\\\&=
4\cdot\dfrac{1}{x}\cdot\dfrac{1}{y^2}
&(\text{use }x^{-m}=\dfrac{1}{x^m})
\\\\&=
\dfrac{4}{xy^2}
.\end{align*}
Hence, the expression $
\dfrac{(4x^{-2})^2(2y^3)}{8x^{-3}y^5}
$ simplifies to $
\dfrac{4}{xy^2}
$.
