\(a^3+6=-3a-2a^2\)
\(\Rightarrow a^3+6+3a+2a^2=0\)
\(\Rightarrow a\left(a^2+3\right)+2\left(a^2+3\right)=0\)
\(\Rightarrow\left(a+2\right)\left(a^2+3\right)=0\)
Vì \(a^2+3>0\forall a\in R\) nên \(a+2=0\Leftrightarrow a=-2\)
\(A=\dfrac{a-1}{a+3}=\dfrac{-2-1}{-2+3}=\dfrac{-3}{1}=-3\)