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Step-by-Step Solution
Step 1: Set up the system of equations
We are given the equations:
$x = cy + bz,$
$y = az + cx,$
$z = bx + ay.$
Rewriting them in a form that sets each to 0, we get:
$-\,x + cy + bz = 0,$
$cx - y + az = 0,$
$bx + ay - z = 0.$
Step 2: Express the system in matrix form
In matrix form, the above system can be written as:
$
\begin{pmatrix}
-1 & c & b \\
c & -1 & a \\
b & a & -1
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
=
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.
$
For a non-trivial (i.e., not all zero) solution $(x, y, z)$ to exist, the determinant of the coefficient matrix must be zero.
Step 3: Compute the determinant and set it to zero
The determinant of the matrix
$
\begin{pmatrix}
-1 & c & b \\
c & -1 & a \\
b & a & -1
\end{pmatrix}
$
must be 0:
$
\Delta
=
\begin{vmatrix}
-1 & c & b \\
c & -1 & a \\
b & a & -1
\end{vmatrix}
= 0.
$
Step 4: Expand the determinant
Using cofactor expansion along the first row (though any row or column works), we get:
$
\Delta
= -1 \bigl((-1)(-1) - a \cdot a \bigr)
\;-\; c \bigl( c(-1) - b \cdot a \bigr)
\;+\; b \bigl( c \cdot a - (-1) \cdot b \bigr).
$
Simplifying each part:
$-1 \left( 1 - a^2 \right) = -1 + a^2.$
$-\,c \left( -\,c - ab \right) = c^2 + abc.$
$b \left( ac + b \right) = abc + b^2.$
Therefore, summing these gives:
$
\Delta = \bigl(-1 + a^2\bigr) + \bigl(c^2 + abc\bigr) + \bigl(abc + b^2\bigr).
$
$
\Delta = -1 + a^2 + b^2 + c^2 + 2abc.
$
Setting $\Delta = 0$ for a non-trivial solution leads to:
$
-1 + a^2 + b^2 + c^2 + 2abc = 0.
$
Step 5: Conclude the required expression
Rewriting the above equation gives:
$
a^2 + b^2 + c^2 + 2abc = 1.
$
This shows that $a^2 + b^2 + c^2 + 2abc$ is equal to 1.
Final Answer
The value of $a^2 + b^2 + c^2 + 2abc$ is $\boxed{1}$.
