$\begin{array}{l}a^3 + b^3 + c^3 - 3abc\\ = (a^3 + b^3) + c^3 - 3abc\\ = (a + b)^3 - 3ab(a + b) + c^3 - 3abc\\ = [(a + b)^3 + c^3] - 3ab(a + b + c)\\ = (a + b + c)[(a + b)^2 - (a + b)c + c^2] - 3ab(a + b + c)\\ = (a + b + c)(a^2 + 2ab + b^2 - ac - bc + c^2) - 3ab(a + b + c)\\ = (a+ b + c)(a^2 + 2ab + b^2 - ac - bc + c^2 - 3ab)\\ = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\\ = 0 \quad (Do \,a^2 +b^2 + c^2 = ab + bc + ca)\\ \Rightarrow a^3 + b^3 + c^3 = 3abc\end{array}$
