If $x = c y + b z, y = c x + a z, z = b x + a y$ the value of $a^{2} + b^{2} - 1$ is

abc

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-abc

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2abc

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-2abc

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Solution

The correct option is D -2abc
$x = 1 y + b z y = c x + a z z = b x + a y x = c y + b (b x + a y) x = c y + b^{2} x + a b y (1 - b^{2}) x = (a b + c) y \frac{x}{y} = \frac{a b + c}{1 - b^{2}} . . . (1) y = c x + a z = c x + a (b x + a y) y = c x + a b x + a^{2} y (1 - a^{2}) y = x (c + a b) \frac{x}{y} = \frac{1 - a^{2}}{c + a b} \frac{y}{x} = \frac{c + a b}{1 - a^{2}} . . . (2)$
Multiplying 1 and 2
$1 = \frac{(a b + c)^{2}}{(1 - a^{2}) (1 - b^{2})} (1 - a^{2}) (1 - b^{2}) = a^{2} b^{2} + c^{2} + 7 a b c 1 - a^{2} - b^{2} + / a^{2} b^{2} = / a^{2} b^{2} + c^{2} + 2 a b c a^{2} + b^{2} + c^{2} + 2 a b c - 1 = 0 a^{2} + b^{2} + c^{2} - 1 = - 2 a b c$

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