Answer
$\dfrac{1}{16}$
Work Step by Step
RECALL:
(i) $a^{-\frac{m}{n}}=\dfrac{1}{a^{\frac{m}{n}}}$
(ii) $a^{\frac{m}{n}}=\sqrt[n]{a^m}$
Use the rules above to obtain:
$=\dfrac{1}{32^{\frac{4}{5}}}
\\=\dfrac{1}{\sqrt[5]{32^4}}
\\=\dfrac{1}{\sqrt[5]{(2^5)^4}}
\\=\dfrac{1}{\sqrt[5]{2^{20}}}
\\=\dfrac{1}{\sqrt[5]{(2^{4})^5}}
\\=\dfrac{1}{2^4}
\\=\dfrac{1}{16}$
