Answer
$\dfrac{a^3}{3^2\sqrt{3}\sqrt[3]{b^7}}$
Work Step by Step
Using the laws of exponents, the given expression, $
3^{-5/2}a^3b^{-7/3}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{a^3}{3^{5/2}b^{7/3}}
.\end{array}
Using $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$, the expression, $
\dfrac{a^3}{3^{5/2}b^{7/3}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{a^3}{\sqrt{3^5}\sqrt[3]{b^7}}
\\\\=
\dfrac{a^3}{\sqrt{3^4\cdot3}\sqrt[3]{b^7}}
\\\\=
\dfrac{a^3}{\sqrt{(3^2)^2\cdot3}\sqrt[3]{b^7}}
\\\\=
\dfrac{a^3}{3^2\sqrt{3}\sqrt[3]{b^7}}
.\end{array}