Let all the expressions be equal to $r$: That way we can solve for $(a,b,c)$ and get $a=b=c=\frac{r\pm\sqrt{r^2+4}}{2}$ These are the only solutions, even over complex numbers. The two expressions in $r$ have range $(0,\infty)$ and $(-\infty,0)$. But $r^2+4\geq 4>0$ so the entire punctured line $a=b=c\neq 0$ is the solution set over reals.

We have $$bc = \frac{c-b}{a-b}$$$$ca = \frac{a-c}{b-c}$$$$ab = \frac{a-b}{a-c}$$ Ergo $(abc)^2 = -1$ implying that there are no real solutions.