The number of triples $(x, y, z)$ of real numbers satisfying the equation $x^{4} + y^{4} + z^{4} + 1 = 4 x y z$ is

0

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4

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8

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more than

8

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Solution

The correct option is B $4$
Using $A . M \geq G . M$
$\frac{x^{4} + y^{4} + z^{4} + 1}{4} \geq (x^{4} . y^{4} . z^{4} .1)^{1 / 4} \Rightarrow x^{4} + y^{4} + z^{4} + 1 \geq 4 x y z$
So, $x^{4} + y^{4} + z^{4} + 1 = 4 x y z$ when
$x^{4} = y^{4} = z^{4} = 1$
We know that $x^{4} + y^{4} + z^{4} + 1 > 0 ∴ x y z > 0$
Hence the possible triples will be
$(1, 1, 1) (- 1, - 1, 1) (- 1, 1, - 1) (1, - 1, - 1)$

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