Answer
Choice I
Work Step by Step
Using the laws of exponents, the given expression, $
\left( n^{\frac{3}{2}}\div n^{-\frac{1}{6}} \right)^{-3}
,$ is equivalent to
\begin{align*}\require{cancel}
&
\left( n^{\frac{3}{2}}\div \dfrac{1}{n^{\frac{1}{6}}} \right)^{-3}
&\left( \text{use }a^{-n}=\dfrac{1}{a^m} \right)
\\\\&=
\left( n^{\frac{3}{2}}\cdot n^{\frac{1}{6}} \right)^{-3}
&\left( \text{multiply by the reciprocal} \right)
\\\\&=
\left( n^{\frac{3}{2}+\frac{1}{6}} \right)^{-3}
&\left( \text{use }a^m\cdot a^n=a^{m+n} \right)
\\\\&=
\left( n^{\frac{9}{6}+\frac{1}{6}} \right)^{-3}
&\left( \text{change to similar fractions} \right)
\\\\&=
\left( n^{\frac{10}{6}} \right)^{-3}
\\\\&=
\left( n^{\frac{\cancel{10}^5}{\cancel6^3}} \right)^{-3}
&\left( \text{divide by $2$} \right)
\\\\&=
\left( n^{\frac{5}{3}} \right)^{-3}
\\\\&=
n^{\frac{5}{3}(-3)}
&\left( \text{use }\left(a^m\right)^n=a^{mn} \right)
\\&=
n^{-5}
.\end{align*}
Hence, Choice I.
