<s>[INST] Fast way to prove that $(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$ What is the most simplest way to prove that $$(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$$
Please!! Thanxx
 [/INST] For brevity's sake, let $x = a - b$, $y = b - c$, and $z = c - a$. We compute a few products:
$$-xz = (a - b)(a - c) = a^2 - ab - ac + bc$$
$$-xy = (b - a)(b - c) = b^2 - ab - bc + ac$$
$$-yz = (c - a)(c - b) = c^2 - ac - bc + ab$$
Adding these all together gives:
$$ - (xy + yz + zx) = a^2 + b^2 + c^2 - ab - ac - bc $$
which we want to square.
$$
\begin{align*}
(a^2 + b^2 + c^2 - ab - ac - bc)^2 &= (xy + yz + xz)^2 \\
&= x^2y^2 + y^2z^2 + x^2z^2 + 2 xyz^2 + 2 xy^2z + 2 x^2yz \\
&= x^2y^2 + y^2z^2 + x^2z^2 + 2 xyz (x + y + z)
\end{align*}
$$
But $x + y + z$ is zero, so we're left with the desired conclusion:
$$
\begin{align*}
(a^2 + b^2 + c^2 - ab - ac - bc)^2 &= x^2y^2 + y^2z^2 + x^2z^2 \\
&= (a - b)^2(b - c)^2 + (b - c)^2(c - a)^2 + (a - b)^2(c - a)^2
\end{align*}
$$
</s>