x³-7x+6=0
(=) x³ + 3x²-3x²-9x+2x+6=0
(=) x²(x+3)-3x(x+3)+2(x+3)=0
(=)(=) (x+3)(x²-3x+2=0)=0
(=) \(\left[ \begin{array}{l}x+3=0\\x²-3x+2=0\end{array} \right.\)
(=) \(\left[ \begin{array}{l}x=-3\\x²-3x+2=0 (*)\end{array} \right.\)
Xét phương trình (*):
x²-3x+2=0
(=) x²-2x-x+2=0
(=) x(x-2)-(x-2)=0
(=) (x-2)(x-1)=0
(=) \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
Vậy x ∈ {-3;2;1}
