Prove that iff the vertices represent an equilateral triangle, then $a^2+\beta^2+\gamma^2-\alpha\beta - \gamma\alpha - \beta\gamma=0$

Question

Show that the distinct complex numbers $\alpha,\beta,\gamma$ represent the vertices of an equilateral triangle (taken in any order) if and only if $$a^2+\beta^2+\gamma^2-\alpha\beta - \gamma\alpha - \beta\gamma=0$$

If $\alpha,\beta,\gamma$ represent the vertices of a triangle, in anticlockwise order then $$(\gamma-\alpha)e^{-\frac{i\pi}{3}}=\beta-\alpha$$ $$(\alpha-\beta)e^{-i\frac{\pi}{3}} = \gamma-\beta$$ $$(\alpha-\gamma)e^{i\frac{\pi}{3}} = \beta-\gamma$$

Multiplying equation 1 and equation 3, then $$(\gamma-\alpha)(\alpha-\gamma)=(\beta-\alpha)(\beta-\gamma)$$ $$\alpha\gamma -\gamma^2 -\alpha^2+\alpha\gamma = \beta^2-\beta\gamma - \alpha\beta +\alpha\gamma$$

$$\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\alpha\gamma=0$$

To complete the iff $\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\alpha\gamma=0$ then we must show this expression implies we have an equilateral triangle.

$$\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\alpha\gamma=0\implies(\gamma-\alpha)(\alpha-\gamma)=(\beta-\alpha)(\beta-\gamma)\implies$$ $$|\gamma-\alpha||\alpha-\gamma|=|\beta-\alpha|\beta-\gamma|$$

$$|\alpha-\gamma|^2=|\beta-\alpha||\beta-\gamma|$$

As the quadratic expression above in $\alpha,\beta,\gamma$ is cyclic then we can write similar expressions for all sides.

$$|\beta-\alpha|^2=|\gamma-\beta||\gamma-\alpha|$$

$$|\gamma-\beta|^2=|\alpha-\gamma||\alpha-\beta|$$

I don't know how to make my proof more obvious than this. But if the square of the sides of a triangle equals the product of the other two lengths, then surely the only way that could happen is if the lengths of the sides are equal?

The triangle that I am referring to in this problem:

Answer 1

Essentially, to finish this argument off, you need to show that, if $a, b, c \ge 0$ such that $a^2 = bc$, $b^2 = ac$, and $c^2 = ab$, then $a = b = c$. A simple way to do this is to order these numbers from least to greatest, i.e. so that $a \le b \le c$, and observe that: $$c = \sqrt{ab} \le \sqrt{b^2} = b \le c,$$ and similarly, $$a = \sqrt{bc} \ge \sqrt{b^2} = b \ge a.$$