<s>[INST] show that $\cos(\theta_{2}-\theta_{3})+\cos(\theta_{3}-\theta_{1})+\cos(\theta_{1}-\theta_{3})+1=0$ If $$a = \cos\theta_{1} + i\sin\theta_{1}, \\ b = \cos\theta_{2} + i\sin\theta_{2}, \\ c = \cos\theta_{3} + i\sin\theta_{3}$$ and $a+b+c=abc$, 
then show that 

$$\cos(\theta_{2}-\theta_{3})+\cos(\theta_{3}-\theta_{1})+\cos(\theta_{1}-\theta_{2})+1=0$$

I am confused with where to start. I did try to proceed with the De Moivre's Theorem but was not able to get hold of the required equation. 
Kindly provide a hint. 
Thanks in advance !
 [/INST] Let $\bar{z}$ denote the complex conjugate of $z$.  Multiplying $a+b+c = abc$ by $\bar{a}$, we get
    \begin{align*}
 1 + \bar{a}b + \bar{a}c = bc
 \end{align*}
    Similarly,
    \begin{align*}
 1+\bar{b}c+\bar{b}a &= ca\\
 1+\bar{c}a + \bar{c}b & = ab
 \end{align*}
    Adding we get,
    \begin{align*}
 3 + (\bar{a}b +a\bar{b}+ \bar{b}c +b\bar{c}+ \bar{c}a + c\bar{a}) = ab+bc+ca
 \end{align*}
    Also, 
    from $a+b+c = abc$, we get $\bar{a} +\bar{b}+\bar{c} = \bar{a}\bar{b}\bar{c}$ and hence dividing throughout by $\bar{a}\bar{b}\bar{c}$ we get
    \begin{align*}
 bc+ca+ab = 1
 \end{align*}
    Thus,
    \begin{align*}
 (\bar{a}b +a\bar{b}+ \bar{b}c +b\bar{c}+ \bar{c}a + c\bar{a}) = -2
 \end{align*}
    and hence 
    \begin{align*}
 \cos(\theta_2-\theta_3) + \cos(\theta_3 - \theta_1) +  \cos(\theta_1 - \theta_2) + 1 = 0 
\end{align*}
</s>