Question

If $x=cy+bz,y=az+cx,z=bx+ay$ where $x,y,z$ are not all zero, then $a^2+b^2+c^2+2abc=k$

Answer 1

We have $x-cy-bz=0cx-y+az=0bx+ay-z=0$

Since $x,y,z$ are not all zero, so the sysmtem will have a non-trivial solution. Therefore,

$\left|\begin{array}{ccc}1& -c& -b\\ c& -1& a\\ b& a& -1\end{array}\right|=0\Rightarrow 1.(1-a^2)+c(-c-ab)-b(ac+b)=0\Rightarrow a^2+b^2+c^2+2abc=1$