Finding $abc$ for distinct reals satisfying $\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$ The problem is

Let $a$, $b$, $c$ be three distinct real numbers satisfying
$$\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$$ Find $abc$.

I can solve this in the case of $a, b, c \ge 0$ by assuming $a =\max\{a, b, c\}$. In that case, $abc =0$. But I haven't solved it yet in the general case.
Please help me. Thanks.

As @fleablood observed, none of $a,b,c$ can be zero. Hence, notice that the condition is equivalent to $$\frac{1+ab}{a}=\frac{1+bc}{b}=\frac{1+ca}{c}\iff \frac1a+b=\frac1b+c=\frac1c+a$$ Yet this implies that $$\frac1a+b=\frac1b+c\iff b-c=\frac{a-b}{ab}$$ Similarly $$\frac1b+c=\frac1c+a\iff c-a=\frac{b-c}{bc}\quad\text{and}\quad  \frac1c+a=\frac1a+b\iff a-b=\frac{c-a}{ca}$$ Multiply the three resulting equations to obtain $$(b-c)(c-a)(a-b)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2}$$ Since $a,b,c$ are distinct real numbers, we have that $a-b\not= 0, b-c\not=0, c-a\not=0$ and thus $$a^2b^2c^2=1\iff abc=\pm1$$ Can you take it from here? Can both $1$ and $-1$ be attained?
