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# Let $a,b,c$ be any real numbers. Suppose that there are real numbers $x,y,z$ not all zero. Such that $x=cy+bz,y=az+cx,z=bx+ay$. Then $a^2+b^2+2abc$ is equal to

$(a)\;2\qquad(b)\;-1\qquad(c)\;0\qquad(d)\;1$

Given equations are
$x-cy-bz=0$
$cx-y+az=0$
$bx+ay-z=0$
$\Rightarrow \begin{vmatrix}1&-c&-b\\c&-1&a\\b&a&-1\end{vmatrix}$
$\Rightarrow 1(1-a^2)+c(-c-ab)-b(ac+b)=0$
$\Rightarrow 1-a^2-c^2-abc-b^2-abc=0$
$\Rightarrow a^2+b^2+c^2+2abc=1$
Hence (d) is the correct answer.
edited Mar 19, 2014