Answer
$81^{-\frac{3}{4}}=\dfrac{1}{27}$
Work Step by Step
RECALL:
(i) $a^{\frac{b}{c}}=\sqrt[c]{a^b}$
(ii) $a^{-m} = \dfrac{1}{a^m}$
Use rule (ii) above to have:
$81^{-\frac{3}{4}}=\dfrac{1}{81^{\frac{3}{4}}}$
Use rule (i) above to have:
$\dfrac{1}{81^{\frac{3}{4}}}
\\=\dfrac{1}{\sqrt[4]{81^3}}
\\=\dfrac{1}{\sqrt[4]{(3^4)^3}}
\\=\dfrac{1}{\sqrt[4]{3^{12}}}
\\=\dfrac{1}{\sqrt[4]{(3^3)^4}}
\\=\dfrac{1}{3^3}
\\=\dfrac{1}{27}$
