Let $a,b,c$ be complex numbers such that $|a|=|b|=|c|=1$. If $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}=1$ as well, then find the product of all possible values of $|a+b+c|$.
Suppose we know that $s^3 = abc(3|s|^2-2),$ where $s = a+b+c.$ Then $|s|$ is a nonnegative root of $x^3 \pm (-3x^2+2)$. For $x^3-3x^2+2=(x-1)(x^2-2x-2),$ the positive roots are $1$ and $1+\sqrt{3}.$ Note that $x^3+3x^2-2 = (x+1)(x^2+2x-2)$ and so it has the single positive real root $-1+\sqrt{3}.$ So the desired product is indeed 2, provided all of the above values are obtainable.
Question: why are all the above values obtainable? I'd prefer an answer that at least provides some motivation as to how to show that both $\pm 1+\sqrt{3}$ are possible values for s.
To show that $s^3 = abc(3|s|^2-2)$, we have the following series of equalities:
$\begin{align}
s^3 &= (a+b+c)^3\\
&= (a+b+c)(a^2 + b^2 + c^2 + 2(ab+ac+bc))\\
&= (a+b+c)(a^2 +b^2 + c^2 – (ab+ac+bc)) + 3(a+b+c)(ab+ac+bc)\\
&= a^3 + b^3 + c^3 – 3abc + 3(a^2 b + ab^2 + abc + a^2 c + abc + ac^2 + abc+b^2 c + bc^2)\\
&= abc-3abc +3abc(a+b+c)(1/a+1/b+1/c)\\
&= abc(-2 + 3|s|^2),
\end{align}$
where the last step follows from the fact that $\overline{s} = 1/a+1/b+1/c$, which in turns follows from the fact that $|a|=|b|=|c|=1$.
Best Answer
As an approach (which almost solves the problem in a different way) to show why the values are obtainable:
Let's assume $a=e^{i \theta_1}, b=e^{i \theta_2}$, and $c=e^{i \theta_3}$. Then, because of $\frac{a^2}{bc}+ \frac{b^2}{ac}+\frac{c^2}{ab}=1$, we must have: $$\cos (2\theta_1-\theta_2-\theta_3)+\cos (2\theta_2-\theta_1-\theta_3)+\cos (2\theta_3-\theta_2-\theta_1)=1 \\ \sin (2\theta_1-\theta_2-\theta_3)+\sin (2\theta_2-\theta_1-\theta_3)+\sin (2\theta_3-\theta_2-\theta_1)=0.$$
Now, set $x=2\theta_1-\theta_2-\theta_3$, $y=2\theta_2-\theta_1-\theta_3$, and $z=2\theta_3-\theta_2-\theta_1$. To make the computations easier, we impose the condition $0\leq y\leq x\leq 2\pi$ (we might lose generality but our goal is to obtain the desired values).
Note that $x+y=-z$. Therefore: $$\sin x +\sin y=\sin (x+y)\\ \cos x+ \cos y=1- \cos (x+y)\\ \implies \\ 2\sin x \sin y+\sin^2 x+\sin^2y=\sin^2(x+y) \\ 2\cos x \cos y+\cos^2 x+\cos^2y=\ 1+cos^2(x+y)-2 \cos (x+y) \\ \implies \\ 2+2\cos (x-y)=2-2\cos (x+y) \\ \implies \\ \cos (x-y)=- \cos (x+y).$$
Hence, Two cases occure.
Case $1$: $\cos (x-y)= \cos (x+y)=0$.
In this case, regarding $0\leq y\leq x\leq 2\pi$, we conclude that $(x,y)=(\frac{\pi}{2}, 0), (2\pi, \frac{\pi}{2}),$ or $(\frac{3\pi}{2}, 0)$; note that not every solution to $\cos (x-y)=- \cos (x+y)$ is acceptable because the major equations are $\sin x +\sin y=\sin (x+y)$ and $\cos x+ \cos y=1- \cos (x+y)$.
Now, if $(x,y)=(\frac{\pi}{2}, 0)$, then $x-y=3(\theta_1-\theta_2)=\frac{\pi}{2}\implies \theta_1-\theta_2 =\frac{\pi}{6}.$
Moreover, $$|a+b+c|^2=(\sin \theta_1+\sin \theta_2+\sin \theta_3)^2+(\cos \theta_1+\cos \theta_2+\cos \theta_3)^2\\ =3+2(cos (\theta_1-\theta_2)+cos (\theta_3-\theta_2)+cos (\theta_1-\theta_3))\\=3+2\cos(\theta_1-\theta_2)+4\cos(\frac{2\theta_3-\theta_2-\theta_1}{2}) \cos(\frac{\theta_1-\theta_2}{2});$$
but $\theta_1-\theta_2=\frac{\pi}{6}$, and $\frac{z}{2}=\frac{2\theta_3-\theta_2-\theta_1}{2}=\frac{-\pi}{4}$. A simple calculation yields $|a+b+c|^2=4+2\sqrt 3.$
Similarly, if $(x,y)=(2\pi,\frac{\pi}{2}),$ it is easy to check that $|a+b+c|^2=1.$
Similarly, if $(x,y)=(\frac{3\pi}{2},0),$ it is easy to check that $|a+b+c|^2=1.$
Case $2$: $x+y=\pi+x-y.$
In this case $y=\frac{\pi}{2}$, and it is not hard to show that $(x,y)=(\frac{3\pi}{2}, \frac{\pi}{2})$ (or $(2\pi, \frac{\pi}{2})$, which is a repeated case).
Similarly, if $(x,y)=(\frac{3\pi}{2}, \frac{\pi}{2})$, then $\theta_1-\theta_2=\frac{\pi}{3}$ and $\frac{z}{2}=\frac{2\theta_3-\theta_2-\theta_1}{2}=-\pi$. So, $|a+b+c|^2=4-2\sqrt 3.$
Notice that $\pi -(x-y)=x+y$ is possible which leads to a repeated case.