Answer
$9^{-\frac{3}{2}}=\dfrac{1}{27}$
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:
$9^{-\frac{3}{2}}=\dfrac{1}{9^{\frac{3}{2}}}$
Use rule (i) above to have:
$\dfrac{1}{9^{\frac{3}{2}}}
\\=\dfrac{1}{\sqrt{9^3}}
\\=\dfrac{1}{\sqrt{(3^2)^3}}
\\=\dfrac{1}{\sqrt{3^6}}
\\=\dfrac{1}{\sqrt{(3^3)^2}}
\\=\dfrac{1}{3^3}
\\=\dfrac{1}{27}$
