a) Khi m = 0, phương trình trở thành:
\(x^3-x^2+x^2-x-6x+6=0\\ \Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)-6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+3x-2x-6\right)=0\\ \Leftrightarrow\left(x-1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=2\end{matrix}\right.\)
Vậy S = {1; -3; 2}