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

We have,

$x^4+4y^4+16z^4+64=32xyz$

We know $AM\ge GM$

$\Rightarrow \frac{x^4+4y^4+16z^4+64}{4}\ge {\left(x^4\times 4y^4\times 16z^4\times 64\right)}^{1/4}$

$\Rightarrow x^4+4y^4+16z^4+64\ge 32xyz$

But given $x^4+4y^4+16z^4+64=32xyz$

$\therefore x^4=4y^4=16z^4=64$

$\Rightarrow$$x=\pm 2\sqrt{2},y=\pm 2,z=\pm \sqrt{2}$

For $x, y, z$
For $x^4+4y^4+16z^4+64=32xyz$

Either each of $x, y, z$ is positive $\to 1$ case or two of $x, y, z$ are negative $\to 3$ cases

$\therefore 4$ cases of different $\left(x, y, z\right)$ triplets $4$ possible $x+y+z$ values $\left(x\ne y\ne z\right).$