The problem is
Suppose for three distinct complex numbers $a, b, c$ such that $|a|=|b|=|c|>0$ all of the three numbers $a+bc, b+ac, c+ab$ are purely real. Prove that $abc=1$
By playing with the identities, I came up with a solution. Let $|a|=|b|=|c|=r$. Then $a+bc$ is real iff:
$$a+bc=\overline{a}+\overline{bc}=\frac{r^2}{abc}\left(bc+r^2a\right)\tag1$$
Thus,
$$\frac{r^2}{abc} = \frac{a^2+abc}{r^2a^2+abc}=\frac{b^2+abc}{abc+r^2b^2}=\frac{c^2+abc}{abc+r^2c^2} \tag2$$
where $(2)$ is obtained by simply solving $(1)$ for $\frac{r^2}{abc}$ and multiplying the denominator and numerator of the fraction obtained by $a$.
Rewriting $(2)$, using the properties of ratios:
$$\frac{r^2}{abc} = \frac{a^2+abc}{r^2a^2+abc}=\frac{b^2+abc}{abc+r^2b^2}=\frac{(a^2+abc)-(b^2+abc)}{(a^2r^2+abc)-(b^2r^2+abc)}=\frac{1}{r^2}$$
which gives $r^4=abc$. Taking the modulus of both sides gives $r^4=r^3$ which implies $r=1$ and so $abc=1$.
I hit upon this solution after an hour or so of trying and even that was by blind luck. Even after solving this, I can't understand the motivation behind the solution and why this particular way of manipulation succeeded.
Is there any intuition behind this solution? Why does this particular method work?
