An algebra problem by Nutan Sinha

Given that $\begin{cases} a+b+c = 1 \\ ab+bc+ca = 2 \\ abc = 3 \end{cases}$ , then by Vieta's formula , $a$ , $b$ , and $c$ are roots of $x^3-x^2+2x-3=0$ . This means that $\begin{cases} a^3-a^2+2a-3=0 \\ b^3-b^2+2b-3=0 \\ c^3-c^2+2c-3=0 \end{cases}$ . Now consider

$\begin{aligned} S & = \sum_{cyc} \frac 1{a+bc} & \small \color{#3D99F6} \text{Multiply up and down by }a \\ & = \sum_{cyc} \frac a{a^2+\color{#3D99F6} abc} & \small \color{#3D99F6} \text{Note that }abc = 3 \\ & = \sum_{cyc} \frac a{a^2+\color{#3D99F6} 3} & \small \color{#3D99F6} \text{From } a^3-a^2+2a-3=0 \implies a^3+2a = a^2+3 \\ & = \sum_{cyc} \frac a{a^3+2a} \\ & = \sum_{cyc} \frac 1{a^2+2} & \small \color{#3D99F6} \text{From } a^3+2a = a^2+3 \implies a(a^2+2) = a^2+2+1 \\ & = \sum_{cyc} (a-1) & \small \color{#3D99F6} \implies a = 1+\frac 1{a^2+2} \\ & = a -1 +b-1+c-1 & \small \color{#3D99F6} \text{Note that }a+b+c = 1 \\ & = 1 -3 = \boxed{-2} \end{aligned}$