Answer
$\color{blue}{2k^{5}}$
Work Step by Step
RECALL:
(1) $a^{-m} = \dfrac{1}{a^m}$
(2) $\dfrac{a^m}{a^n} = a^{m-n}$
(3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$
(4) $(ab)^m = a^mb^m$
(5) $(a^m)^n=a^{mn}$
(6) $a^m \cdot a^n = a^{m+n}$
(7) $a^0=1, a\ne0$
Use rule (5) above to obtain:
$=\dfrac{12k^{-2}(k^{-3\cdot (-4)})}{6k^5}
\\=\dfrac{12k^{-2}(k^{12})}{6k^5}$
Use rule (6) above to obtain:
$=\dfrac{12k^{-2+12}}{6k^5}
\\=\dfrac{12k^{10}}{6k^5}$
Divide the coefficients by cancelling common factors to obtain:
$\require{cancel}
\\=\dfrac{\cancel{12}2k^{10}}{\cancel{6}k^5}
\\=\dfrac{k^{10}}{k^5}$
Use rule (2) above to obtain:
$=2k^{10-5}
\\=\color{blue}{2k^{5}}$