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If $x^2+y^2+z^2 \neq 0, \quad x=c y+b z, \quad y=a z+c x$ and $z=b x+a y$, then $a^2+b^2+c^2+2 a b c$ is equal to
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Verified Answer
The correct answer is:
$1$
We have,
$$
\begin{aligned}
x-c y-b z & =0 \\
-c x+y-a z & =0 \\
-b x-a y+z & =0
\end{aligned}
$$
Eliminating $x, y, z$, we get
$$
\begin{array}{rr}
& \left|\begin{array}{ccc}
1 & -c & -b \\
-c & 1 & -a \\
-b & -a & 1
\end{array}\right|=0 \\
\Rightarrow & 1\left(1-a^2\right)+c(-c-a b)-b(c a+b)=0 \\
\Rightarrow & 1-a^2-c^2-a b c-a b c-b^2=0 \\
\Rightarrow & a^2+b^2+c^2+2 a b c=1
\end{array}
$$
$$
\begin{aligned}
x-c y-b z & =0 \\
-c x+y-a z & =0 \\
-b x-a y+z & =0
\end{aligned}
$$
Eliminating $x, y, z$, we get
$$
\begin{array}{rr}
& \left|\begin{array}{ccc}
1 & -c & -b \\
-c & 1 & -a \\
-b & -a & 1
\end{array}\right|=0 \\
\Rightarrow & 1\left(1-a^2\right)+c(-c-a b)-b(c a+b)=0 \\
\Rightarrow & 1-a^2-c^2-a b c-a b c-b^2=0 \\
\Rightarrow & a^2+b^2+c^2+2 a b c=1
\end{array}
$$
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