Let $z_1, z_2$ and $z_3$ be complex numbers such that $z_1+z_2+z_3=2, z_1^2+z_2^2+z_3^2=3$ and $z_1z_2z_3=4$. Find $\frac1{z_1z_2+z_3-1}+\frac1{z_2z_3+z_1-1}+\frac1{z_3z_1+z_2-1}$
My Attempt:
I got $\sum z_2z_3=\frac12$
Therefore, the cubic equation whose roots are $z_1,z_2,z_3$ is $2x^3-4x^2+x-8=0$. Let this equation be $(1)$.
Also, $\frac1{z_1z_2+z_3-1}=\frac{z_3}{4+z_3^2-z_3}$
Therefore, we need to find the sum of roots of the cubic equation whose roots are of the form $\frac{z_3}{4+z_3^2-z_3}$
Let $x=\frac{z_3}{4+z_3^2-z_3}$
From this, $z_3$ comes out to be $\frac{(x+1)\pm\sqrt{(x+1)^2-16x}}{2}$
But $z_3$ is a root of equation $(1)$. So, maybe I should put the value of $z_3$ here and obtain the cubic in $x$.
But is there any other approach for this question?
I had also tried taking LCM in the required expression but couldn't finish.
Best Answer
Note that: $$\sum_{cyc} \frac{1}{ab+c-1} = \sum_{cyc} \frac{1}{ab+c-(a+b+c-1)} = \sum_{cyc} \frac{1}{(a-1)(b-1)}$$ So we have to evaluate a decently easy value, it becomes:
$$\frac{a+b+c-3}{(a-1)(b-1)(c-1)}$$ where the denominator is easily expanded as a cubic in $-1$. $-1 +1(a+b+c) -1(ab+bc+ca) + abc$
So the final expression should be $\frac{-2}{9}$.