Answer
$\dfrac{1}{16}$
Work Step by Step
RECALL:
(i) $a^{\frac{m}{n}}=(\sqrt[n]{a})^m.$
(ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0.$
Use rule (ii) above to obtain
$\dfrac{1}{32^{\frac{4}{5}}}.$
Use rule (i) with m=4 and n=5 to obtain
$\dfrac{1}{(\sqrt[5]{32})^4}
\\=\dfrac{1}{(\sqrt[5]{2^5})^4}
\\=\dfrac{1}{(2)^4}
\\=\dfrac{1}{16}.$
