Answer
$\dfrac{27\sqrt 2}{32} $
Work Step by Step
Use the rule $a^{-m}=\left(\dfrac{1}{a}\right)^m$ to obtain:
$\left (\frac{8}{9} \right )^{-\frac{3}{2}}=\left (\frac{9}{8} \right )^{\frac{3}{2}}$
Use $9=3^2$ and $8=2^3$.
$=\left (\dfrac{3^2}{2^3} \right )^{\frac{3}{2}}$
Use the rule $\left(\dfrac{a}{b}\right)^n=\dfrac{a^m}{b^m}$:
$=\dfrac{(3^2)^{\frac{3}{2}}}{(2^3)^{\frac{3}{2}}} $
Use the rule $\left(a^m\right)^n=a^{mn}$, then simplify to obtain:
$=\dfrac{3^{2\cdot \frac{3}{2}}}{2^{3\cdot \frac{3}{2}}} $
$=\dfrac{3^{3}}{2^{\frac{9}{2}}} $
$=\dfrac{27}{2^{\frac{9}{2}}} $
Multiply the numerator and the denominator by $2^{\frac{1}{2}}$.
$=\dfrac{27}{2^{\frac{9}{2}}} \cdot \dfrac{2^{\frac{1}{2}}}{2^{\frac{1}{2}}}$
Use the rule $a^m \cdot a^n = a^{m+n}$, then simplify to obtain:
$=\dfrac{27\sqrt2}{2^{\frac{9}{2}+\frac{1}{2}}} $
$=\dfrac{27\sqrt2}{2^{5}} $
$=\dfrac{27\sqrt2}{32} $
Hence, the correct answer is $\frac{27\sqrt 2}{32} $.
