Let a,b,c be any real numbers.Suppose that there are real numbers x,y,z not all zero such that $x = c y + b z, y = a z + c x$ and $z = b x + a y$ , then $a^{2} + b^{2} + c^{2} + 2 a b c$ is equal to

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Solution

The correct option is D 1
$a, b, c$ real numbers

$x, y, z$ real numbers not all zero

$x = c y + b z \to x - c y - b z = 0 - - - (1)$

$y = a z + c x \to c x + y - a z = 0 - - - (2)$

$z = b x + c y \to - b x - a y + z = 0 - - - (3)$

The system of equation have trivial solution then

$∣ ∣ ∣ ∣ \begin{matrix} 1 & - c & - b c & - 1 & a b & a & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow 1 (1 - a^{2}) + c (- c - a b) - b (a + b) = 0$

$\Rightarrow a^{2} + b^{2} + c^{2} + 2 a b c = 1$

$D$ is correct

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