10. (20 points) Given a sequence $\left\{a_{n}\right\}$ whose terms are all non-zero real numbers, and for any positive integer $n$ we have
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3} \text {. }
$$
(1) When $n=3$, find all sequences of three terms $a_{1}, a_{2}, a_{3}$ that satisfy the condition.
(2) Does there exist an infinite sequence $\left\{a_{n}\right\}$ that satisfies the condition and has $a_{2013}=-2012$? If it exists, find a general term formula for such an infinite sequence; if not, explain the reason.
10. (1) When $n=1$, $a_{1}^{2}=a_{1}^{3}$.

By $a_{1} \neq 0$, we get $a_{1}=1$.
When $n=2$, $\left(1+a_{2}\right)^{2}=1+a_{2}^{3}$.
By $a_{2} \neq 0$, we get $a_{2}=2$ or -1.
When $n=3$, $\left(1+a_{2}+a_{3}\right)^{2}=1+a_{2}^{3}+a_{3}^{3}$.
If $a_{2}=2$, we get $a_{3}=3$ or -2;
If $a_{2}=-1$, we get $a_{3}=1$.
In summary, there are three sequences of three terms that satisfy the conditions:
$1,2,3$ or $1,2,-2$ or $1,-1,1$.
(2) Let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Then
$S_{n}^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}\left(n \in \mathbf{N}_{+}\right)$.
Thus, $\left(S_{n}+a_{n+1}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n+1}^{3}$.
Subtracting the two equations and combining with $a_{n+1} \neq 0$, we get
$$
2 S_{n}=a_{n+1}^{2}-a_{n+1} \text {. }
$$

When $n=1$, from (1) we know $a_{1}=1$;
When $n \geqslant 2$,
$$
\begin{array}{l}
2 a_{n}=2\left(S_{n}-S_{n-1}\right) \\
=\left(a_{n+1}^{2}-a_{n+1}\right)-\left(a_{n}^{2}-a_{n}\right),
\end{array}
$$

which simplifies to $\left(a_{n+1}+a_{n}\right)\left(a_{n+1}-a_{n}-1\right)=0$.
Therefore, $a_{n+1}=-a_{n}$ or $a_{n}+1$.
Given $a_{1}=1, a_{2013}=-2012$, then
$$
a_{n}=\left\{\begin{array}{ll}
n, & 1 \leqslant n \leqslant 2012 ; \\
(-1)^{n} 2012, & n \geqslant 2013 .
\end{array}\right.
$$