<s>[INST] If $\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then $|z_{1}+z_{2}+z_{3}|$ 
If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$
and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then possible values of $|z_{1}+z_{2}+z_{3}|$

$\bf{My\; Try::}$ Given $$\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1\Rightarrow z^3_{1}+z_{2}^3+z_{3}^3=-z_{1}z_{2}z_{3}$$
Now Using $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
So $$\left(z_{1}+z_{2}+z_{3}\right)\left[z^2_{1}+z^2_{2}+z^2_{3}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}\right] = -4z_{1}z_{2}z_{3}$$
So $$(z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)$$
Now Let $k=z_{1}+z_{2}+z_{3}.$ Taking Conjugate, we get $\displaystyle k = \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = \frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}z_{2}z_{3}}$
So we get $z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=kz_{1}z_{2}z_{3}$
Put into $(1)\;,$ We get $$(z_{1}+z_{2}+z_{3})\cdot \left[(z_{1}+z_{2}+z_{3})^2-3kz_{1}z_{2}z_{3}\right] = -4z_{1}z_{2}z_{3}$$
Now How can I solve after that, Help Required, Thanks
 [/INST] Here is a 3rd solution (inspired by yours, @Christian Blatter, but using different variables). 
Let  $$Z_1=\frac{z^2_{1}}{z_{2}z_{3}}, \ Z_2=\frac{z^2_{2}}{z_{3}z_{1}}, Z_3=\frac{z^2_{3}}{z_{1}z_{2}} \ \ \ (*)$$
Let us remark that all $Z_k \in \mathbb{U}$ (unit circle).
Thus the initial condition is replaced by $Z_1+Z_2+Z_3=-1 \ \ \ (1)$
Another immediate relationship is $Z_1Z_2Z_3=1 \ \ \ (2)$
At last, is we set $w=Z_1Z_2+Z_2Z_3+Z_3Z_1$, we have
$$w=\frac{z_{1}z_{2}}{z^2_{3}}+\frac{z_{3}z_{1}}{z^2_{2}}+\frac{z_{2}z_{3}}{z^2_{1}}$$
Using the rule $u \in\mathbb{U} \Longrightarrow \bar{w}=\dfrac{1}{w}$ :
$\bar{w}=Z_1+Z_2+Z_3$ thus $w=-1 \ \ \ (3) \ $ (by using (1))
Taking into account (1), (2), (3), by Vieta's formulas, $\{Z_1,Z_2,Z_3\}$ is the set of roots of 
$$Z^3+Z^2-Z-1=0$$
i.e., $1$ (multiplicity 1) and $-1$ (multiplicity 2).
One can assume, using definitions (*), that $Z_1=1,Z_2=Z_3=-1$ ; thus
$$\begin{cases}z_1^2=z_2z_3\\ z_2^2=-z_3z_1\\ z_3^2=-z_1z_2\end{cases}  \ \ \ (4)$$
or, equivalently:
$$\dfrac{z_1}{z_2}=\dfrac{z_3}{z_1}=-\dfrac{z_2}{z_3}=:\zeta \ \ \ (5)$$
(i.e., we have given the name $\zeta$ to this common ratio).
Thus, by elementary operations on (5), one finds 
$$\zeta^3=-1 \ \ \  \Longrightarrow \ \ \zeta=-1, e^{i \pi/3}, e^{-i \pi/3} \ \ \ (6)$$
Thus, using (5), the set of solutions of the initial equation is  
$$\begin{cases}z_1=z_1\\ z_2=\zeta^2 z_1\\ z_3=\zeta z_1\end{cases} \ \ \ (7)$$
where $z_1$ is any element of $\mathbb{U}$.
Thus $|z_1+z_2+z_3|=|z_1||1+\zeta+\zeta^2|=2|z_1|=2$ whatever the value of $\zeta$.
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