<s>[INST] If $\{a,b,c\}\subset\mathbb C$ such that $a^2+b^2+c^2=ab +bc +ca$ prove that they lie on the vertices of an equilateral triangle in the complex plane 
Exercise 1.11 (from Friendly Approach to Complex Analysis).
If $a$, $b$, $c$ are real numbers such that $a^2+b^2+c^2=ab+bc+ca$, then they must be equal. Indeed, doubling both sides and rearranging gives $(a-b)^2+(b-c)^2+(c-a)^2=0$, and since each summand is nonnegative, it must be that case that each is $0$. On the other hand, now show that if $a$, $b$, $c$ are complex numbers such that $a^2+b^2+c^2=ab+bc+ca$, then they must lie on the vertices of an equilateral triangle in the complex plane. Explain the real case result in the light of this fact.
Hint: Calculate $\big((b-a)\omega+(b-c)\big)\big((b-a)\omega^2+(b-c)\big)$, where $\omega$ is a nonreal cube root of unity.

It is a solved exercise but I don't understand the solution (or the hint). More than that I don't get why the author shared this exercise.
I understand the cube roots of unity are $\{ (1,0), (-\frac{1}{2}, \frac{\sqrt{3}}{2}), (-\frac{1}{2}, -\frac{\sqrt{3}}{2}) \}$
So I have $a := (1,0)$ and $b := (-\frac{1}{2}, \frac{\sqrt{3}}{2}) $ and solved for $c$, and got two values. One of them was the last cube root of unity. However I have observed that the points (solutions for $c$ considered one at a time) lie on the vertices of an equilateral triangle.
Isn't this enough? What is the thought process behind the hint?
 [/INST] Hint.
We have $(b-a)^2+(c-a)^2 = (b-a)(c-a)$ so
$$
c-a = (b-a)\left(\frac{1\pm\sqrt{3}i}{2}\right)
$$
then
$|b-a| = |c-a|$ and $\angle(b-a,c-a) = \pm\frac{\pi}{3}$
NOTE
The property $a^2+b^2+c^2 - a b- a c- b c = 0$ is translation invariant.
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