Answer
$-\dfrac{1}{3a^3}$
Work Step by Step
Using the laws of exponents, the given expression, $
\dfrac{-9a^5}{27a^8}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{-\cancel{9}a^{5-8}}{\cancel{9}(3)}
\\\\=
\dfrac{-(a)^{-3}}{3}
\\\\=
\dfrac{-1}{3a^3}
\\\\=
-\dfrac{1}{3a^3}
.\end{array}