Ta có:
x².(x-1) - 2x(x-3) - 9(x-1) = 0
⇔ (x-1)(x²-9) -2x(x-3) = 0
⇔ (x-1)(x-3)(x+3) - 2x(x-3) = 0
⇔ (x-3)(x² + 2x - 3 - 2x) = 0
⇔ (x-3)(x²-3) = 0
⇔ \(\left[ \begin{array}{l}x-3=0\\x³-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=3\\x=±√3\end{array} \right.\)