This problem is Exercise 5.5.30 of "The Art and Craft of Problem Solving" by Paul Zeitz.
The problem asks to use the identity $ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc) $ to prove the AMGM inequality. I have done this by noting that the right side is equal to $(a+b+c)\frac{1}{2}((a-b)^2 + (b-c)^2 + (a-c)^2)$ and then substituting $a=x^{1/3}$, $b=y^{1/3}$, $c=z^{1/3}$.
The problem then asks "Does this method generalize?", and I'm having trouble finding one. Is there a factorization of $x_1^n + x_2^n + \dots x_n^n - nx_1x_2\cdots x_n$ that can prove AMGM for $n$ variables? Or perhaps using some other generalization of the given identity?