`***`Lời giải`***`
`x^3+(x-1)^3=(2x-1)^3`
`<=>x^3+x^3-3x^2+3x-1=8x^3-12x^2+6x-1`
`<=>(x^3+x^3-8x^3)+(12x^2-3x^2)+(3x-6x)=-1+1`
`<=>-6x^3+9x^2-3x=0`
`<=>-3x(2x^2-3x+1)=0`
`<=>`\(\left[ \begin{array}{l}-3x=0\\2x^2-3x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\2x^2-2x-x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\2x(x-1)-(x-1)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\(x-1)(2x-1)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x-1=0\\2x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=1\\2x=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=1\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `S={0;1;1/2}`
