1+2+x1=x3+812
⇔1+2+x1−x3+812=0
⇔1+x+21−(x+2)(x2−2x+4)12=0
⇔(x+2)(x2−2x+4)+(x2−2x+4)−12=0
⇔x3−2x2+4x+2x2−4x+8+x2−2x+4−12=0
⇔x3+x2−2x=0
⇔x(x2+x−2)=0
⇔x(x2+x−1−1)=0
⇔x[(x2−1)+(x−1)]=0
⇔x[(x−1)(x+1)+(x−1)]=0
⇔x(x−1)(x+2)=0
⇔⎣⎡x=0x−1=0x+2=0⇔⎣⎡x=0x=1x=−2
⇒S={0;1;−2}