\(x^3-3x^2-x+3=0\)
\(x^2\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right)\left(x^2-1\right)=0\)
\(\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)
x3-3x2-x+3=0
⇒x(x2-1)-3(x2-1)=0
⇒(x-3)(x2-1)=0
⇒(x-3)(x-1)(x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0+3=3\\x=0+1=1\\x=0-1=-1\end{matrix}\right.\)
Vậy \(x\in\left\{3;\pm1\right\}\)
x\(^3\)-3x\(^2\)-x+3
=(x\(^3\)-3x\(^2\))-(x-3)
=x\(^2\)(x-3)-(x-3)
=(x-3)(x\(^2\)-1)
