2016/078)if $x=cy+bz$, $y= az + cx$ and $z=bx+ay$ then show that $a^2+b^2+c^2 + 2abc = 1$

we put the equations in standard form
$x-cy - bz = 0\cdots(1)$                    
$cx -y + az = 0\cdots(2)$    
$bx + ay - z = 0\cdots(3)$                            
multiply (1) by c and subtract it from (2) to get
$(c^2-1) y + (bc+ a) z = 0\cdots(4)$
mulltiply (1) by b and subtract from (3) to get
$(a+bc) y - (1-b^2) z = 0\cdots(5)$
as (4) and (5) are consistant we get
$(c^2-1)(b^2 - 1) -   (bc+ a)^2 = 0$
or $b^2c^2 - b^2 - c^2 + 1 -  b^2c^2 - a^2 - 2bca$
or $a^2+b^2+c^2 + 2abc = 1$