Answer
$\dfrac{r^{5}}{81s^{10}}$
Work Step by Step
Using the laws of exponents, the given expression, $
\dfrac{(3r^{-2}s^3t^0)^{-3}}{3rs}
,$ is equivalent to
\begin{align*}
&
\dfrac{ 3^{1(-3)}r^{-2(-3)}s^{3(-3)}t^{0(-3)}}{3rs}
&\text{ (use $\left(a^x\right)^y=a^{xy})$}
\\\\&=
\dfrac{3^{-3}r^{6}s^{-9}t^{0}}{3rs}
\\\\&=
\dfrac{3^{-3}r^{6}s^{-9}(1)}{3rs}
&\text{ (use $a^0=1$)}
\\\\&=
\dfrac{3^{-3}r^{6}s^{-9}}{3rs}
\\\\&=
3^{-3-1}r^{6-1}s^{-9-1}
&\text{ $\left(\text{use } \dfrac{a^x}{a^y}=a^{x-y}\right)$}
\\\\&=
3^{-4}r^{5}s^{-10}
\\\\&=
\dfrac{r^{5}}{3^{4}s^{10}}
&\text{ $\left(\text{use } a^{-x}=\dfrac{1}{a^x}\right)$}
\\\\&=
\dfrac{r^{5}}{81s^{10}}
.\end{align*}
Hence, the simplified form of the given expression is $
\dfrac{r^{5}}{81s^{10}}
$.
