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(C) Given the equation $x^4+4y^4+16z^4+64=32xyz$.Applying the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality to the four positive terms $x^4

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$4$

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(C) Given the equation $x^4+4y^4+16z^4+64=32xyz$.Applying the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality to the four positive terms $x^4, 4y^4, 16z^4, 64$:$\frac{x^4+4y^4+16z^4+64}{4} \geq \sqrt[4]{x^4 \cdot 4y^4 \cdot 16z^4 \cdot 64} = \sqrt[4]{4096 x^4 y^4 z^4} = 8|xyz|$.Thus,$x^4+4y^4+16z^4+64 \geq 32|xyz|$.For the equality to hold,we must have $x^4 = 4y^4 = 16z^4 = 64$,which implies $|x|=2\sqrt{2}, |y|=2, |z|=\sqrt{2}$.Also,the product $xyz$ must be positive for the equality $32xyz = 32|xyz|$ to hold.The possible combinations for $(x, y, z)$ are:$1$. $(2\sqrt{2}, 2, \sqrt{2}) \implies x+y+z = 2\sqrt{2}+2+\sqrt{2} = 3\sqrt{2}+2$.$2$. $(2\sqrt{2}, -2, -\sqrt{2}) \implies x+y+z = 2\sqrt{2}-2-\sqrt{2} = \sqrt{2}-2$.$3$. $(-2\sqrt{2}, 2, -\sqrt{2}) \implies x+y+z = -2\sqrt{2}+2-\sqrt{2} = 2-3\sqrt{2}$.$4$. $(-2\sqrt{2}, -2, \sqrt{2}) \implies x+y+z = -2\sqrt{2}-2+\sqrt{2} = -\sqrt{2}-2$.There are $4$ distinct possible values for the sum $x+y+z$.

The number of different possible values for the sum $x+y+z$,where $x, y, z$ are real numbers such that $x^4+4y^4+16z^4+64=32xyz$ is

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