You can subtitute a=1, b=1, or c=1

Relevant wiki: Vieta's Formula Problem Solving - Basic

$\begin{aligned} \chi & = \frac {ab+b+{\color{#3D99F6}1}}{ca+a+1} + \frac {ca+a+{\color{#3D99F6}1}}{bc+c+1} + \frac {bc+c+{\color{#3D99F6}1}}{ab+b+1} & \small \color{#3D99F6} \text{Vieta's formula: }abc = 1 \\ & = \frac {ab+b+{\color{#3D99F6}abc}}{ca+a+1} + \frac {ca+a+{\color{#3D99F6}abc}}{bc+c+1} + \frac {bc+c+{\color{#3D99F6}abc}}{ab+b+1} \\ & = \frac {b\cancel{(a+1+ca)}}{\cancel{ca+a+1}} + \frac {a\cancel{(c+1+bc)}}{\cancel{bc+c+1}} + \frac {c\cancel{(b+1+ab)}}{\cancel{ab+b+1}} \\ & = \color{#3D99F6}b + a + c & \small \color{#3D99F6} \text{Vieta's formula: }a+b+c = 2017 \\ & = \boxed{\color{#3D99F6}2017} \end{aligned}$

From Vieta's Theorem, we get $a+b+c=2017$ and $abc = 1$ And then, let $a=\frac{p}{q}$ , $b=\frac{q}{r}$ , and $c=\frac{r}{p}$ . Starting from $\frac{ab+b+1}{ca+a+1}+\frac{ca+a+1}{bc+c+1}+\frac{bc+c+1}{ab+b+1}$ $=\frac{\frac{p}{q}.\frac{q}{r}+\frac{q}{r}+1}{\frac{r}{p}.\frac{p}{q}+\frac{p}{q}+1}+\frac{\frac{r}{p}.\frac{p}{q}+\frac{p}{q}+1}{\frac{q}{r}.\frac{r}{p}+\frac{r}{p}+1}+\frac{\frac{q}{r}.\frac{r}{p}+\frac{r}{p}+1}{\frac{p}{q}.\frac{q}{r}+\frac{q}{r}+1}$ $=\frac{\frac{p}{r}+\frac{q}{r}+1}{\frac{r}{q}+\frac{p}{q}+1}+\frac{\frac{r}{q}+\frac{p}{q}+1}{\frac{q}{p}+\frac{r}{p}+1}+\frac{\frac{q}{p}+\frac{r}{p}+1}{\frac{p}{r}+\frac{q}{r}+1}$ $=\frac{\frac{p+q+r}{r}}{\frac{p+q+r}{q}}+\frac{\frac{p+q+r}{q}}{\frac{p+q+r}{p}}+\frac{\frac{p+q+r}{p}}{\frac{p+q+r}{r}}$ $=\frac{q}{r}+\frac{p}{q}+\frac{r}{p}$ $=a+b+c$ $=\boxed{2017}$