Answer
$\dfrac{4k^{17}}{125}$
Work Step by Step
Using the laws of exponents, the given expression, $
\dfrac{(2k)^2k^3}{k^{-1}k^{-5}}(5k^{-2})^{-3}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4k^2k^3(5^{-3}k^{-2(-3)})}{k^{-1}k^{-5}}
\\\\=
\dfrac{4k^{2+3}(5^{-3}k^{6})}{k^{-1+(-5)}}
\\\\=
\dfrac{4k^{5}(k^{6})}{5^{3}k^{-6}}
\\\\=
\dfrac{4k^{5+6-(-6)}}{5^{3}}
\\\\=
\dfrac{4k^{5+6+6}}{125}
\\\\=
\dfrac{4k^{17}}{125}
.\end{array}