Ta có : a^3+b^3+c^3=3abc⇔a^3+b^3+c^3−3abc=0
⇔(a+b)^3+c^3−3ab(a+b)−3abc=0
⇔(a+b+c)(a^2+b^2+2ab−bc−ac)−3ab(a+b+c)=0
⇔(a+b+c)(a^2+b^2+c^2−ab−bc−ac)=0
⇔a+b+c^2[(a^2−2ab+b^2)+(b^2−2bc+c^2)+(c^2−2ac+a^2)]=0
⇔a+b+c^2[(a−b)^2+(b−c)^2+(c−a)^2]=0
⇔[a+b+c=0 ; (a−b)^2+(b−c)^2+(c−a)^2=0
⇔[a+b+c=0 ; a=b=c