Three, (25 points) For a positive integer $k$, find the greatest common divisor $d$ of all positive even integers $x$ that satisfy the following conditions:
(1) $\frac{x+2}{k}$ and $\frac{x}{k}$ are both integers, and the difference in the number of digits of these two numbers equals the difference between these two numbers;
(2) The product of the digits of $\frac{x}{k}$ is a perfect cube.
From (1) we know that $\frac{x+2}{k}-\frac{x}{k}=\frac{2}{k}$ is an integer.
Since $k$ is a positive integer, we have $k=1$ or 2.
When $k=1$, the number of digits in $x+2$ is $(x+2)-x=2$ more than the number of digits in $x$, which is clearly impossible.

When $k=2$, the number of digits in $\frac{x+2}{2}$ is $\frac{x+2}{2}-\frac{x}{2}=1$ more than the number of digits in $\frac{x}{2}$.

Thus, the digits of $\frac{x}{2}$ are all 9, otherwise, the number of digits in $\frac{x+2}{2}$ would be the same as that in $\frac{x}{2}$.

Let the number of digits in $\frac{x}{2}$ be $n$, and the product of its digits be $m^{3}$ (where $m$ is a positive integer).
Then $9^{n}=m^{3} \Rightarrow 3^{2 n}=m^{3} \Rightarrow 312 n$.
Since $(3,2)=1$, we have $31 n$.
Let $n=3 t$ (where $t$ is a positive integer). Then
$$
\begin{array}{l}
\left.\frac{x}{2}=\underset{n \uparrow}{99 \ldots 9}=\underset{34 \uparrow}{99 \ldots 9} \Rightarrow 999 \right\rvert\, \frac{x}{2} \\
\Rightarrow 1998 \mid x \Rightarrow 1998 \leqslant d .
\end{array}
$$

When $t=1$, $x=1998$.
And $d \leqslant x \leqslant 1998$, so $d=1998$.