<s>[INST] Prove that $(a^2-bc)^3+(b^2-ac)^3+(c^2-ab)^3=3(a^2-bc)(b^2-ac)(c^2-ab)$ 
Let $a+b+c=0$. Prove that 
  $$(a^2-bc)^3+(b^2-ac)^3+(c^2-ab)^3=3(a^2-bc)(b^2-ac)(c^2-ab)$$

My work so far:
I used the formula
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
Then $$(a^2-bc)^3+(b^2-ac)^3+(c^2-ab)^3-3(a^2-bc)(b^2-ac)(c^2-ab)=$$
$$=(a^2+b^2+c^2-ab-bc-ca)\times$$
$$\times\left((a^2-bc)^2+(b^2-ac)^2+(c^2-ab)^2-(a^2-bc)(b^2-ac)-(b^2-ac)(c^2-ab)-(a^2-bc)(c^2-ab) \right)$$
But if $a\not=b$ or $a\not=b$ or $a\not=c$ then $a^2+b^2+c^2-ab-bc-ca>0$.
How do I prove 
$$(a^2-bc)^2+(b^2-ac)^2+(c^2-ab)^2-(a^2-bc)(b^2-ac)-(b^2-ac)(c^2-ab)-(a^2-bc)(c^2-ab)=0?$$
 [/INST] Let $a^2-bc=x$, $b^2-ca=y$, $c^2-ab=z$. Then note $$x-y=a^2-bc+ca-b^2=(a-b)(a+b+c)=0$$$$y-z=b^2-ca+ab-c^2=(b-c)(a+b+c)=0$$Thus we have $x=y=z$. We are done. 
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