Answer
$3a^2+4a+4$
Work Step by Step
The given expression, $
\dfrac{3(a+1)^{-1}+4a^{-2}}{(a^3+a^2)^{-1}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{3}{a+1}+\dfrac{4}{a^2}}{\dfrac{1}{a^3+a^2}}
\\\\=
\dfrac{\dfrac{3(a^2)+4(a+1)}{a^2(a+1)}}{\dfrac{1}{a^2(a+1)}}
\\\\=
\dfrac{\dfrac{3(a^2)+4(a+1)}{\cancel{a^2(a+1)}}}{\dfrac{1}{\cancel{a^2(a+1)}}}
\\\\=
\dfrac{3(a^2)+4(a+1)}{1}
\\\\=
3(a^2)+4(a+1)
\\\\=
3a^2+4a+4
.\end{array}
