\(y^3+2y^2-3y\)
\(=y^3+3y^2-y^2-3y\)
\(=y^2\left(y+3\right)-y\left(y+3\right)\)
\(=\left(y^2-y\right)\left(y+3\right)=\left(ay^2+by+c\right)\left(y+3\right)\)
\(\Leftrightarrow y^2-y=ay^2+by+c\)
\(\Leftrightarrow\hept{\begin{cases}a=1\\b=-1\\c=0\end{cases}}\)