[Solution] First, we prove that $x$ is a four-digit number. In fact,
from
$$
44 x-86868>0 \text {, }
$$

we know
$$
x \geqslant\left[\frac{86868}{44}\right]+1=1975 .
$$

Therefore, $x$ is at least a four-digit number.
If the number of digits of $x$, $k \geqslant 5$, then
$$
P(x)=44 x-86868>4 \cdot 10^{k}-10^{5}>3 \cdot 10^{k}>9^{k} \geqslant P(x) .
$$

where $P(x)$ represents the product of the digits of $x$:
The contradiction in the above equation indicates that $x$ is exactly a four-digit number.
Let the sum of the digits of $x$ be $S(x)$, then
$$
1 \leqslant S(x) \leqslant 36 \text {. }
$$

Since $S(x)$ is a perfect cube, then $S(x)=1,8$ or 27.
Clearly, $S(x)=1$ does not meet the requirement.
Also, since $\quad 0<P(x) \leqslant 9^{4}=6561$,
$$
\begin{array}{c}
44 x-86868 \leqslant 6561, \\
x \leqslant\left[\frac{86868+6561}{44}\right]=2123 .
\end{array}
$$

Thus, we have

and
$$
\begin{array}{c}
1975 \leqslant x \leqslant 2123 . \\
S(x)=8 \text { or } 27, \\
P(x) \neq 0 .
\end{array}
$$

It is not difficult to find that such $x$ only has four: $1989,1998,2114,2123$.
By calculating $P(x)$ and $44 x-86868$ for each, we find that only $x=1989$ meets the requirement.
10.65 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 perfect cube.
(52nd Moscow Mathematical Olympiad, 1989)