Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number. 
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.


I wanted to calculate $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:
$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.

$$x = \frac{(a+1)(b+1)(c+1)-(1+abc)}{1+abc} = \frac{(a+1)(b+1)(c+1)}{1+abc}-1$$
and using the property $a \bar{a} = |a|^2 = 1$ repeatedly:
$$\frac{(a+1)(b+1)(c+1)}{1+abc} = \frac{(1+1/a)(1+1/b)(1+1/c)}{1/abc+1} = \frac{(1+\bar{a})(1+\bar{b})(1+\bar{c})}{\overline{abc}+1}$$
so $x = \bar{x}$ (why?) and hence $x$ is real.
