<s>[INST] 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.
 [/INST] 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 $ad + bc + b^2 + c^2 =0$
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