Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

You can make a nice, somewhat abstract argument that connects this; in particular, equilateral triangles have various nice properties - like:

• For fixed $z_1,z_2$, there are exactly two $z_3$ such that $(z_1,z_2,z_3)$ is an equilateral triangle. If $z_1$ and $z_2$ are real, then those $z_3$ are reflections around the real axis.
• Any permutation of an equilateral triangle is an equilateral triangle - so $(z_1,z_2,z_3)$ being equilateral implies $(z_1,z_3,z_2)$ and $(z_2,z_3,z_1)$ are also equilateral.
• The map $z\mapsto az+b$ preserves equilateral triangles.

If we make a leap to trying to think about an analogous polynomial $P(z_1,z_2,z_3)$ such that, at any root, $(z_1,z_2,z_3)$ is an equilateral triangle, then the above conditions translate to:

• $P(z_1,z_2,z_3)$ has degree two in each of its variables. It has only real coefficients.
• $P(z_1,z_2,z_3)$ is a symmetric polynomial.
• $P(z_1,z_2,z_3)$ is homogenous.

Together, these yield that if such a $P$ existed, $P$ would have to be a sum of the only two symmetric and homogenous polynomials of degree two, meaning, for some $\alpha$ and $\beta$: $$P(z_1,z_2,z_3)=\alpha(z_1^2+z_2^2+z_3^2)+\beta(z_1z_2+z_2z_3+z_3z_1)$$ but, since we desired $P(z,z,z)=0$ to be a solution, we can calculate that clearly $\alpha=-\beta$ and, dividing out by $\alpha$ would give that the only polynomial that could feasibly represent equilateral triangles would be: $$P(z_1,z_2,z_3)=z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1.$$ Then, if you showed that if $P(z_1,z_2,z_3)=0$ implied $P(z_1+c,z_2+c,z_3+c)=0$, which could be done algebraically, or perhaps through some clever manipulation, you would have that the roots of this polynomial satisfied all the conditions we wanted of equilateral triangles - and those conditions suffice to uniquely define the set of equilateral triangles, which proves that $P$'s roots are all equilateral triangles.

The point of this answer is, of course, to draw a connection between the symmetries defining the original problem and how the carry over to symmetries in the algebra and, indeed, suffice to show the statement. You could do this the other way round to learn geometric symmetries from the algebraic form of $P$ as well.

Here's an example of what I mean in my comment. The fact that all sides are equal, and that the angles between them are equal can be expressed quite elegantly as $$\frac{z_3-z_2}{z_2-z_1}=\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3}.$$

The desired equation now follows from some elementary algebra. I believe this result can be generalized to any regular n-gon.

If $z_1=w_1+x$, $z_2=w_2+x$ and $z_3=w_3+x$, the identity becomes $$ w_1^2+2w_1x+x^2+ w_2^2+2w_2x+x^2+ w_3^2+2w_3x+x^2=\\ w_1w_2+w_1x+w_2x+x^2+ w_2w_3+w_2x+w_3x+x^2+ w_3w_1+w_3x+w_1x+x^2 $$ that becomes $$ w_1^2+w_2^2+w_3^2=w_1w_2+w_2w_3+w_3w_1 $$ so, taking $x=-z_3$, we can assume $z_3=0$ and we need to prove that if $z_1$, $z_2$ and $0$ are the vertices of an equilateral triangle, then $$ z_1^2+z_2^2=z_1z_2. $$

If $z_1=w_1u$ and $z_2=w_2u$ (with $u\ne0$), the identity above becomes $w_1^2+w_2^2=w_1w_2$, so we can assume $z_1=e^{i\pi/6}$, which means that $z_2=e^{i(\pi/6+\pi/3)}$ or $z_2=e^{i(\pi/6-\pi/3)}$.

If $z_2=e^{i(\pi/6+\pi/3)}=e^{i\pi/2}=i$, then a rotation brings us in the situation with $z_1=e^{-i\pi/6}$ and $z_2=e^{i\pi/6}$. So we are reduced to proving that $$ e^{i\pi/3}+e^{-i\pi/3}=1 $$ which is obvious.