Answer
$\dfrac{m^4}{n}$
Work Step by Step
RECALL:
(i) $a^m \cdot a^n = a^{m+n}$
(ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$
(iii) $a^0=1, a\ne0$
Use the rules above to have:
$=m^{3-(-1)}n^{2-3}
\\=m^{3+1}n^{-1}
\\=m^4 \cdot \dfrac{1}{n}
\\=\dfrac{m^4}{n}$
