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