`(3x+1)^3-(2x-1)^3=(x+2)^3`
`<=> [(3x+1)-(2x-1)][(3x+1)^2+(3x+1)(2x-1)+(2x-1)^2]=(x+2)^3`
`<=> (3x+1-2x+1)(9x^2+6x+1+6x^2-3x+2x-1+4x^2-4x+1)-(x-2)^3=0`
`<=> (x+2)(19x^2+x+1)-(x+2)^3=0`
`<=> (x+2)[19x^2+x+1-(x+2)^2]=0`
`<=>(x+2)(19x^2+x+1-x^2-4x-4)=0`
`<=> (x+2)(18x^2-3x-3)=0`
`<=> 3(x+2)(6x^2-x-1)=0`
`<=> 3(x+2)(6x^2-3x+2x-1)=0`
`<=> 3(x+2)[3x(2x-1)+(2x-1)]=0`
`<=> 3(x+2)(2x-1)(3x+1)=0`
`<=>`\(\left[ \begin{array}{l}x+2=0\\2x-1=0\\3x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{2}\\x=\dfrac{-1}{3}\end{array} \right.\)
Vậy `S={-2;1/2;-1/3}`
