Let's determine those three-digit numbers in the decimal system that are equal to the sum of the cubes of their digits.
Let the digits of the desired number be denoted by $A, B, C$, then

$$
\begin{gathered}
1 \leq A \leq 9,0 \leq B, C \leq 9 \\
100 A+10 B+C=A^{3}+B^{3}+C^{3}, \text { that is } \\
100 A-A^{3}=\left(B^{3}-10 B\right)+\left(C^{3}-C\right)
\end{gathered}
$$

The left side can take 9 values, the terms on the right side can take 10 - 10 values, from which we need to select one each so that (1) is satisfied. The values are:

| $A, B, C=$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| $100 A-A^{3}=a=-$ | 99 | 192 | 273 | 336 | 375 | 384 | 357 | 288 | 171 |  |
| $B^{3}-10 B=b=$ | 0 | $-10+1$ | $-20+8$ | $-10+7$ | 24 | 75 | 156 | 273 | $(432)$ | $(639)$ |
| $C^{3}-C=c=$ | 0 | 0 | 6 | 24 | 60 | 120 | 210 | 336 | $(504)$ | $(720)$ |

The following two observations reduce the trial work. The largest value of $a$ is 384, the largest absolute negative value of $b$ is $-12$, and $c$ is non-negative, so the last two values of $b$ and $c$ cannot appear in the solution. The unit digit of $a$ and $b$ takes all different values (for $B=1,2,3$ we wrote this as positive), while the unit digit of $c$ takes only the values $0,4,6$.

Based on these, for most of the $a, b, c$ triples, we can already state from the unit digit of the numbers that they are not suitable, or instead of discarding, we can perform the selection in a targeted manner. Specifically, if the last digit of $c$ is 0, then the last digit of $a$ and $b$ can only be the same; if the last digit of $c$ is 4 (or 6), then the last digit of $a$ is either 4 greater (or 6 greater) or 6 less (or 4 less) than the last digit of $b$. By checking the total sums of the remaining triples to be examined, we find the following 4 solutions:

| Last digit of $c$: | 0 | 0 | 4 | 6 |
| :--- | ---: | ---: | ---: | ---: |
| Last digit of $b$: | 3 | 3 | 5 | 0 |
| Last digit of $a$: | 3 | 3 | 9 | 6 |
| Value of $c$: | 0 | 0 | 24 | 336 |
| Value of $b$: | 273 | 273 | 75 | 0 |
| Value of $a$: | 273 | 273 | 99 | 336 |
| Value of $C$: | 0 | 1 | 3 | 7 |
| Value of $B$: | 7 | 7 | 5 | 0 |
| Value of $A$: | 3 | 3 | 1 | 4 |
| The desired number: | 370 | 371 | 153 | 407 |