Then find the value of ... Let $a, \: b, \: c$ be three variables which can take any value (Real or Complex). Given that $ab + bc + ca = \frac{1}{2}$; $a + b + c = 2$; $abc = 4$. Then find the value of $$\frac{1}{ab + c - 1} + \frac{1}{bc + a - 1} + \frac{1}{ac + b - 1}$$
I always get stuck in such problems. Please give a detailed solution and also give me an approach that I can use to solve other such kind of problems.

Hint:
Let $y=\dfrac1{ab+c-1}$
$\implies y=\dfrac 1{\dfrac12--bc-ca+c-1}=\dfrac2{2c(1-a-b)-1}=\dfrac2{2c(c-1)-1}=\dfrac2{2c^2-2c-1} $
$$\iff2y c^2-2y c-(y+2)=0\  \ \ \  (1)$$
Again $y=\dfrac1{ab+c-1}=\dfrac c{4+c^2-c}$
$$\iff c^2y-c(y+1)+4y=0\  \ \ \  (2)$$
Solve $(1),(2)$ for $c,c^2$ and use $c^2=(c)^2$ to form a cubic equation in $y$
Now apply Vieta's formula
