hay \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Ta lại có \(a^3+b^3+c^3=3abc\) khi \(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)