solving system of equations with constant $i$

Question

What are the values of $a,b,c$ given the system of equations given below:
$a+b+ab=i$
$b+c+bc=2i$
$c+a+ac=3i$

Answer 1

Suppose that the equations be $$a+b+a\,b=X\tag 1$$ $$b+c+b\,c=Y\tag 2$$ $$c+a+c\,a=Z\tag 3$$ Eliminate $b$ from $(1)$ and $c$ from $(3)$; this gives $$b=\frac{X-a}{a+1}\qquad c=\frac{Z-a}{a+1}$$ Replace these expressions in $(2)$ and simplify; this leads to $$ (Y+1)\,a^2+2 ( Y+1)\,a+(-X Z-X+Y-Z)=0$$ which is a quadratic in $a$.

Replacing $X=i$, $Y=2i$, $Z=3i$ and solving the quadratic which becomes $$a^2+2a -\left(\frac{1}{5}+\frac{8 i}{5}\right)=0 $$ gives $$a_1=-1+2 \sqrt{\frac{2}{5}}+i\sqrt{\frac{2}{5}}\qquad a_2=-1-2 \sqrt{\frac{2}{5}}-i\sqrt{\frac{2}{5}}$$ $$b_1=-1+\frac{3}{\sqrt{10}}+i\frac{1}{\sqrt{10}}\qquad b_2=-1-\frac{3}{\sqrt{10}}-i\frac{1}{\sqrt{10}}$$ $$c_1=-1+\sqrt{\frac{5}{2}}+i\sqrt{\frac{5}{2}}\qquad c_2=-1-\sqrt{\frac{5}{2}}-i\sqrt{\frac{5}{2}}$$

Answer 2

Note that adding $1$ to each equation gives that

$(a+1)(b+1)=i+1$.
$(b+1)(c+1)=2i+1$
$(c+1)(a+1)=3i+1$.

Multiply these gives us that $((a+1)(b+1)(c+1))^2=(i+1)(2i+1)(3i+1)=-10$.

This implies that $(a+1)(b+1)(c+1)=\pm \sqrt{10}i$.

Proceed with Calculations from here.