If $y=\frac {a+bz}{c+dz}$, $z=\frac{a+bx}{c+dx}$, $x=\frac{a+by}{c+dy}$, then $ad + bc + b^2 + c^2 = 0$ I need to solve this problem and I don’t know how.

If $x, y, z$ are unequal and $y = \frac {a + bz}{c + dz}, z = \frac {a + bx}{c + dx}, x = \frac {a + by}{c + dy},$ then $ad + bc + b^2 + c^2 = 0$.

I see that I need to eliminate $x$, $y$, $z$ but I don’t know how to start.
I appreciate your help.
 A: $$
\begin{bmatrix}
y\\ 
1
\end{bmatrix}
=
\alpha
\begin{bmatrix}
b & a\\ 
d & c
\end{bmatrix}
\begin{bmatrix}
z\\ 
1
\end{bmatrix}, \qquad
\begin{bmatrix}
z\\ 
1
\end{bmatrix}
=
\beta
\begin{bmatrix}
b & a\\ 
d & c
\end{bmatrix}
\begin{bmatrix}
x\\ 
1
\end{bmatrix},  \qquad
\begin{bmatrix}
x\\ 
1
\end{bmatrix}
=
\gamma
\begin{bmatrix}
b & a\\ 
d & c
\end{bmatrix}
\begin{bmatrix}
y\\ 
1
\end{bmatrix}
$$
$$
\Rightarrow 
\begin{bmatrix}
y\\ 
1
\end{bmatrix}
=
\alpha\beta\gamma
\begin{bmatrix}
b & a\\ 
d & c
\end{bmatrix}^{3}
\begin{bmatrix}
y\\ 
1
\end{bmatrix}, \qquad
\begin{bmatrix}
z\\ 
1
\end{bmatrix}
=
\alpha\beta\gamma
\begin{bmatrix}
b & a\\ 
d & c
\end{bmatrix}^{3}
\begin{bmatrix}
z\\ 
1
\end{bmatrix},  \qquad
\begin{bmatrix}
x\\ 
1
\end{bmatrix}
=
\alpha\beta\gamma
\begin{bmatrix}
b & a\\ 
d & c
\end{bmatrix}^{3}
\begin{bmatrix}
x\\ 
1
\end{bmatrix}
$$
Since $x$, $y$, $z$ are different, hence this is only possible iff
$$
\alpha\beta\gamma
\begin{bmatrix}
b & a\\ 
d & c
\end{bmatrix}^{3}
=
\begin{bmatrix}
1 & 0\\ 
0 & 1
\end{bmatrix}
$$
$$
\alpha\beta\gamma
\begin{bmatrix}
b^{3}+ad(2b+c) & a(b^{2}+c^{2} +ad+bc) \\ 
d(b^{2}+c^{2} +ad+bc) & c^{3}+ad(b+2c)
\end{bmatrix}
=
\begin{bmatrix}
1 & 0\\ 
0 & 1
\end{bmatrix}
$$
so either {$a=0$ and $d=0$} or $b^{2}+c^{2} +ad+bc =0$
A: Substitute formula for x into formula for z we have
$$z = \frac{ac + ady + ab + b^2y}{c^2 + cdy + ad + bdy}$$
Substitute this into formula for y we have
$$(c^3 + acd)y + (c^2d + bcd + ad^2 + b^2d)y^2 = (ac^2 + a^2d + abc + ab^2) + (bad + b^3)y$$
$$(c^3 + acd - abd - b^3)y + d(c^2 + bc + ad + b^2)y^2 = a(ad + bc + b^2 + c^2)$$
$$y(ad(c - b) + (c^3 - b^3)) + d(ad + bc + b^2 + c^2)y^2 = a(ad + bc + b^2 + c^2)$$
$$(ad + bc + b^2 + c^2)(dy^2 + (c - b)y - a) =0$$
Hence $ad + bc + b^2 + c^2 =0$
