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
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3} .
$$
(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 $a_{2}=-1$.

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 $a_{3}=-2$; if $a_{2}=-1$, we get $a_{3}=1$.

In summary, there are 3 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 $S_{n}^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}(n \in \mathbf{N}^{*}$.
Thus $\left(S_{n}+a_{n+1}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}+a_{n+1}^{3}$.
Subtracting the two equations and combining with $a_{n+1} \neq 0$, we get $2 S_{n}=a_{n+1}^{2}-a_{n+1}$.
When $n=1$, from (1) we know $a_{1}=1$;
When $n \geqslant 2$, $2 a_{n}=2\left(S_{n}-S_{n-1}\right)=\left(a_{n+1}^{2}-a_{n+1}\right)-\left(a_{n}^{2}-a_{n}\right)$, i.e., $\left(a_{n+1}+a_{n}\right)\left(a_{n+1}-a_{n}-1\right)=0$, so $a_{n+1}=-a_{n}$ or $a_{n+1}=a_{n}+1$.
Given $a_{1}=1, a_{2013}=-2012$, there exists a form that satisfies the conditions:
$$
a_{n}=\left\{\begin{array}{ll}
n, & 1 \leqslant n \leqslant 2012, \\
2012(-1)^{n}, & n \geqslant 2013
\end{array}\right.
$$