\(a^3+b^3+c^3=3abc\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc=3abc\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\left(vô.lí\right)\end{matrix}\right.\)
Vậy a+b+c=0