Đáp án:
`(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0`
`<=>(x^2-1)[(x^2-1)^2-(x^4+x^2+1)]=0`
`<=>(x^2-1)(x^4-2x^2+1-x^4-x^2-1)=0`
`<=>-3x^2(x^2-1)=0`
`<=>x^2(x^2-1)=0`
`<=>` \(\left[ \begin{array}{l}x^2=0\\x^2-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x^2=0\\x^2=1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\)
Vậy `S={0,1,-1}`.
