`(x-3)^3+(x+1)^3=8(x-1)^3`
`<=>x^3-9x^2+27x-27+x^3+3x^2+3x+1=8(x^3-3x^2+3x-1)`
`<=>2x^3-6x^2+30x-26=8x^3-24x^2+24x-8`
`<=>2x^3-8x^3-6x^2+24x^2+30x-24x-26+8=0`
`<=>-6x^3+18x^2+6x-18=0`
`<=>-6(x^3-3x^2-x+3)=0`
`<=>(x^3-x)-(3x^2-3)=0`
`<=>x(x^2-1)-3(x^2-1)=0`
`<=>(x-3)(x^2-1)=0`
`<=>(x-3)(x-1)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x-3=0\\x-1=0\\x+1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=3\\x=1\\x=-1\end{array} \right.\)
Vậy `S={-1;1;3}`
