10. 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.
(1) When $n=3$, $a_{1}^{2}=a_{1}^{3} \Rightarrow a_{1}=1,\left(a_{1}+a_{2}\right)^{2}=a_{1}^{3}+a_{2}^{3} \Rightarrow\left(1+a_{2}\right)^{2}=1+a_{2}^{3}$ $\Rightarrow a_{2}^{2}-a_{2}-2=0 \Rightarrow a_{2}=2$ or -1.
If $a_{2}=2,\left(a_{1}+a_{2}+a_{3}\right)^{2}=a_{1}^{3}+a_{2}^{3}+a_{3}^{3} \Rightarrow\left(3+a_{3}\right)^{2}=9+a_{3}^{3}$
$\Rightarrow a_{3}^{2}-a_{3}-6=0 \Rightarrow a_{3}=3$ or -2;
If $a_{2}=-1,\left(a_{1}+a_{2}+a_{3}\right)^{2}=a_{1}^{3}+a_{2}^{3}+a_{3}^{3} \Rightarrow a_{3}^{2}=a_{3}^{3} \Rightarrow a_{3}=1$.
Therefore, the three-term sequences $a_{1}, a_{2}, a_{3}$ that satisfy the conditions are three: $1,2,3 ; 1,2,-2 ; 1,-1,1$.
(2) When $a_{k}=k(k=1,2, \cdots, n)$, $\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}$.
$$
\begin{array}{l}
\left(a_{1}+a_{2}+\cdots+a_{n}+a_{n+1}\right)^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}+a_{n+1}^{3} \\
\Rightarrow\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}+2\left(a_{1}+a_{2}+\cdots+a_{n}\right) a_{n+1}+a_{n+1}^{2} \\
=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}+n(n+1) a_{n+1}+a_{n+1}^{2}=a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}+a_{n+1}^{3} \\
\Rightarrow a_{n+1}^{2}-a_{n+1}-n(n+1)=0 \Rightarrow a_{n+1}=n+1 \text { or }-n .
\end{array}
$$

Thus, taking $a_{k}=k(k=1,2, \cdots, 2012) \Rightarrow a_{2013}=-2012$.
At this point, $\left(a_{1}+a_{2}+\cdots+a_{2013}\right)^{2}=(1+2+\cdots+2011)^{2}=1^{3}+2^{3}+\cdots+2011^{3}$,
Thus, $\left(a_{1}+a_{2}+\cdots+a_{2013}+a_{2014}\right)^{2}=\left(1+2+\cdots+2011+a_{2014}\right)^{2}$
$=(1+2+\cdots+2011)^{2}+2011 \cdot 2012 a_{2014}+a_{2014}^{2}=1^{3}+2^{3}+\cdots+2011^{3}+a_{2014}^{3}$
$\Rightarrow a_{2014}^{2}-a_{2014}-2011 \cdot 2012=0 \Rightarrow a_{2014}=2012$ or -2011.
Thus, the subsequent terms can be $2012,-2012,2012,-2012, \cdots$;
or $-2011,-2010, \cdots,-2,-1,1,-1,1,-1 \cdots$;
or $2012,2013,2014, \cdots$; and after any term, the above three situations can be repeated. Therefore, a general term formula for the infinite sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=\left\{\begin{array}{ll}
n, & n=1,2, \cdots, 2012 \\
2012 \cdot(-1)^{n}, & n=2013,2014, \cdots .
\end{array}\right.
$$