Show that $(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$ is a perfect square and find its square root.
My work:
Let, $x^2-yz=a,y^2-zx=b,z^2-xy=c$. So, we can have,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=\dfrac12[(x-y)^2+(y-z)^2+(z-x)^2]\cdot\dfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$
Now, I got into a mess. I have got two products with sum of three squares which I cannot manage nor can I show this to be a perfect square. Please help.
HINT:
$$(x^2-yz)^3-(x^2-yz)(y^2-zx)(z^2-xy)$$
$$=(x^2-yz)[(x^2-yz)^2-(y^2-zx)(z^2-xy)]$$
$$=(x^2-yz)x(x^3+y^3+z^3-3xyz)$$
$$=(x^3-xyz)\underbrace{(x^3+y^3+z^3-3xyz)}$$
As the terms under the brace is symmetric wrt $x,y,z$
we shall reach at the similar expressions from
$$(y^2-zx)^3-(x^2-yz)(y^2-zx)(z^2-xy)$$ and $$(z^2-xy)^3-(x^2-yz)(y^2-zx)(z^2-xy)$$
Use the formula $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=\frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$ and then notice that $a-b=x^2-y^2+zx-yz=(x-y)(x+y+z)$ This will lead you to get the answer as $(x^3+y^3+z^3-3xyz)^2$ as pointed out by Ewan Delanoy.
