Answer
$\dfrac{1}{cd^4}$
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:
$=c^{3-4}d^{-5-(-1)}
\\=c^{-1}d^{-5+1}
\\=\dfrac{1}{c} \cdot d^{-4}
\\=\dfrac{1}{c} \cdot \dfrac{1}{d^4}
\\=\dfrac{1}{cd^4}$
