Cyclic polynomial proof

If$$x+y+z = 0$$ Then prove, $$(x^2+xy+y^2)^3+(y^2+yz+z^2)^3+(z^2+zx+x^2)^3$$ $$=3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)$$

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HINT:

Observe that if we put $x^2+xy+y^2=a$ etc., we need to prove $a^3+b^3+c^3=3abc$

From this, the above proposition will be true

either if $a+b+c=0$

or if $a=b=c$ for real $a,b,c$

Now, $x^2+xy+y^2-(y^2+yz+z^2)=x^2-z^2+xy-yz=(x-z)(x+z+y)=0$

Alternatively eliminating $x,$

$x^2+xy+y^2=x(x+y)+z^2=\{-(y+z)\}(-z)+y^2$ as $x+y+z=0$

$\implies x^2+xy+y^2=y^2+yz+z^2$

Similarly we can prove, $x^2+xy+y^2=z^2+zx+x^2$

$$\implies x^2+xy+y^2=y^2+yz+z^2=z^2+zx+x^2$$

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