\(\left(x-1\right)\left(x^2-x+1\right)=x^3-1\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+1\right)=\left(x-1\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\left(x^2-x+1\right)-\left(x^2+x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+1-x^2-x-1\right)=0\)
\(\Leftrightarrow-2x\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2x=0\\x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(\left(x-1\right)\left(x^2-x+1\right)=x^3-1\)
=> \(\left(x-1\right)\left(x^2-x+1\right)=(x-1)\left(x^2+x+1\right)\)
=> \(\left(x-1\right)\left(x^2-x+1\right)=(x-1)\left(x^2-x+1+2x\right)\)
=> \(\left(x-1\right)\left(x^2-x+1\right)=(x-1)\left(x^2-x+1\right)+\left(x-1\right).2x\)
=> 0 = \(\left(x-1\right).2x\)
=> \(\left[{}\begin{matrix}x-1=0\\2x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1\right\}\)
\(=x^3-x^2+x-x^2+x-1=x^3-1\)
\(=-x^2+x-x^2+x=0\)
\(=-2x^2+2x=0\)
\(=-2x\left(x-1\right)=0\)
\(=x\left(x-1\right)=0\)
\(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x_1=0;x_2=1\)
