W6 Find all natural numbers $x$ that satisfy the following conditions: the product of the digits of $x$ equals $44x - 86868$, and the sum of the digits is a cube number.
Given $44 x \geqslant 86868$, hence $x \geqslant 1975$, so $x$ must be at least a four-digit number. But if the number of digits $k \geqslant 5$, then
$$
44 x-86868>4 \times 10^{k}-10^{5}>3 \times 10^{k}>9^{k} \geqslant P(x),
$$

where $P(x)$ denotes the product of the $k$ digits of $x$, leading to a contradiction. Therefore, $x$ must be a four-digit number.
Now, the sum of the digits of $x$, $S(x)$, satisfies $1 \leqslant S(x) \leqslant 36$. Since $S(x)$ is a perfect cube, $S(x)=1$, 8, or 27. Clearly, $S(x)=1$ is not feasible. Also, because $0<P(x) \leqslant 9^{4}=6561$, we have
$$
x \leqslant\left[\frac{86868+6561}{44}\right]=2123 .
$$

In summary, we know that $1975 \leqslant x \leqslant 2123$, and $S(x)=8$ or $27$, $P(x) \neq 0$. It is not difficult to find that such $x$ only has four possibilities: $1989, 1998, 2114, 2123$. Upon inspection, only $x=1989$ meets the requirements.