Equilateral triangles and complex numbers

Factorisation. With $\omega^2+\omega+1=0$ and $\omega^3=1$,
$$(a+\omega b+\omega^2c)(a+\omega^2b+\omega c)=a^2+b^2+c^2-ab-bc-ca.$$
So the condition holds iff $a+\omega b+\omega^2c=0$ or $a+\omega^2b+\omega c=0$.

Geometric meaning. Since $1+\omega+\omega^2=0$, we have $a+\omega b+\omega^2c=(a-c)+\omega(b-c)$. Hence $a+\omega b+\omega^2c=0$ iff
$$a-c=-\omega\,(b-c)=e^{-i\pi/3}(b-c):$$
the vector $\vec{CA}$ is $\vec{CB}$ rotated by $-60^\circ$, so $ABC$ is equilateral. Likewise $a+\omega^2b+\omega c=0$ gives the other orientation.

Conclusion. $a^2+b^2+c^2=ab+bc+ca$ iff $a,b,c$ form an equilateral triangle (degenerate when $b=c$ forces $a=b=c$).

Lesson. Factor symmetric expressions with cube roots of unity: multiplication by $\omega$ is a rotation by $120^\circ$.