3. Find all positive integers $n$ such that the equation
$$
x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}
$$

has positive integer solutions.
(Suppose $x \leqslant y \leqslant z$. Clearly, $z^{2} \mid \left(x^{3}+y^{3}\right)$, then $z^{2} \leqslant x^{3}+y^{3}$. It is easy to see that $z=n x^{2} y^{2}-\frac{x^{3}+y^{3}}{z^{2}} \geqslant n x^{2} y^{2}-(x+y)$. Therefore, $x^{3}+y^{3} \geqslant z^{2} \geqslant\left[n x^{2} y^{2}-(x+y)\right]^{2}$. Thus, we have $n^{2} x^{4} y^{4} \leqslant 2 n x^{2} y^{2}(x+y)+x^{3}+y^{3}-(x+y)^{2}$, which means $x y < 2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}}$. It is easy to see that $x=1$, then $y<2+\frac{2}{y}+\frac{1}{n}+\frac{1}{n y^{3}}$. Hence, $y \leqslant 3$. Since $z^{2} \mid \left(1+y^{3}\right)$, it can only be $y=1$, $z=1$, or $y=2, z=3$. Therefore, $n=1$ or 3.)