$x^3$-$x^2$-$6x$=$0$
⇔ $x$($x^2$-$x$-$6$)=$0$
⇔ $x$($x^2$-$3x$+$2x$-$6$)=$0$
⇔ $x$($x-3$)($x+2$)=$0$
⇔ \(\left[ \begin{array}{l}x=0\\x-3=0\\x+2=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=0\\x=3\\x=-2\end{array} \right.\)
⇒ $S$=${0;-2;3}$
Xin hay nhất !