Is It the same problem type?

Let $\displaystyle x^3+y^3+z^3=nx^2y^2z^2$ . We claim that $n$ can only be $1,3$ , which makes the answer $\boxed {4}$ .

WLOG $x\le y\le z$ , then $3z^3\ge nx^2y^2z^2\implies z\ge \dfrac {nx^2y^2}{3}$ . In addition, $z^2|x^3+y^3+z^3\implies z^2|x^3+y^3\implies$ $x^3+y^3\ge z^2\ge \dfrac {n^2}{9}x^4y^4$ . Thus $2y^3\ge x^3+y^3\ge \dfrac {n^2}{9}x^4y^4\implies$ $\frac {18}{n^2}\ge x^4y\ge 1$

Hence $n$ can only take values $1,2,3,4$ .

When $n=4$ , $(x,y)=(1,1)$ . So the equation becomes: $z^3-4z^2+2=0$ . By the rational root test \ $z\in {1,2}$ , both of which fail.

When $n=3$ , take the solution $(x,y,z)=(1,1,1)$ .

When $n=2$ , $(x,y)=(1,1), (1,2)$ . By plugging these into the original equation and applying the rational root test, we find that these are not valid possibilities.

When $n=1$ , take $(x,y,z)=(1,2,3)$ .

In conclusion, $1,3$ are the only possible values for $n$ .