Answer
$\color{blue}{\dfrac{1}{16}}$
Work Step by Step
RECALL:
(1) $a^{m/n} = \left(a^{1/n}\right)^m$
(2) $a^{1/n} = \sqrt[n]{a}$
(3) For positive real numbers $a$, $\sqrt[n]{a^n}=a$
(4) $a^{-m} = \dfrac{1}{a^m}$
(5) When $n$ is odd, $\sqrt[n]{a^n} = n$
Use rule (4) above to obtain:
$(-32)^{-4/5} = \dfrac{1}{(-32)^{4/5}}$
Use rule (1) above to obtain:
$=\dfrac{1}{[(-32)^{1/5}]^4}$
Use rule (2) above to obtain:
$=\dfrac{1}{(\sqrt[5]{-32})^4}$
With $-32=(-2)^5$, the expression above is equivalent to:
$=\dfrac{1}{\left(\sqrt[5]{(-2)^5}\right)^4}$
Use rule (5) above to obtain:
$=\dfrac{1}{(-2)^4}
\\=\color{blue}{\dfrac{1}{16}}$
