`\text{~~Holi~~}`
`x^3-1=-x(x-1)`
`-> (x-1)(x^2+x+1)=-x(x-1)`
`-> (x-1)(x^2+x+1)+x(x-1)=0`
`-> (x-1)(x^2+x+1+x)=0`
`-> (x-1)(x^2+2x+1)=0`
`->`\(\left[ \begin{array}{l}x-1=0\\x^2+2x+1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `S={±1}`
`=>` Đáp án `D.`Có 2 nghiệm phân biệt