8.93 Find all arithmetic sequences $a_{1}, a_{2}, \cdots, a_{n}$, such that each term is an integer, and
$$
\begin{array}{l}
a_{1}^{3}+a_{2}^{3}+a_{3}^{3}=a_{4}^{3}, \\
a_{1}+a_{2}+\cdots+a_{n}=500 .
\end{array}
$$
[Solution] Let the common difference of the sequence be $d$. From the given conditions, we have
$$
\begin{array}{l}
a_{1}^{3}+\left(a_{1}+d\right)^{3}+\left(a_{1}+2 d\right)^{3}=\left(a_{1}+3 d\right)^{3}, \\
n a_{1}+\frac{n(n-1)}{2} d=500,
\end{array}
$$

where $a_{1}, d$ are integers, and $n$ is a natural number with $n \geqslant 4$. From (1), after simple calculations, we get
$$
a_{1}^{3}=6 a_{1} d^{2}+9 d^{3},
$$

which implies $\quad\left(a_{1}-3 d\right)\left(a_{1}^{2}+3 d a_{1}+3 d^{2}\right)=0$.
If $a_{1}^{2}+3 d a_{1}+3 d^{2}=0$, then $a_{1}=d=0$, which contradicts (2)! Therefore, we must have
$$
a_{1}=3 d \text {. }
$$

Substituting (3) into (2) yields
$$
\left(3 n+\frac{1}{2} n(n-1)\right) d=500 \text {. }
$$

This implies that $3 n+\frac{1}{2} n(n-1)$ is a divisor of 500, where $n \geqslant 4$ is a natural number. For any natural number $m$, if there exists a natural number $n \geqslant 4$ such that $3 n+\frac{1}{2} n(n-1)=m$, i.e.,
$$
n^{2}+5 n-2 m=0 \text {, }
$$

then $25+8 m$ must be a perfect square, and $25+8 m \geqslant 13^{2}=169$, i.e., $m \geqslant 18$. Furthermore, since $m$ is a divisor of 500, simple calculations show that $m=25$ or 250. From (4) and (3), we have
$$
n=5, a_{1}=60, d=20 \text { or } n=20, a_{1}=6, d=2 \text {. }
$$

After verification, the following sequences
$$
\begin{array}{l}
60,80,100,120,140, \\
6,8,10,12, \cdots, 44,
\end{array}
$$

are the only solutions.