Some relations between multi-trace combinations I recently checked a relation between three $2 \times 2$ matrices $\mathrm{A,B,C}$.
$$\text{tr (ABC)}+\text{tr (ACB)}+\text{tr(A)tr(B)tr(C)}=\text{tr(A)tr(BC)}+\text{tr(B)tr(AC)}+\text{tr(C)tr(AB)}$$
Each component of the above relation makes all the possible multi-trace combinations. I checked a similar relation between four $3\times 3$ matrices is also possible. I believe some smart guy had found such relations long times before even more generally. Could you tell me any related theories if you know about it? Thank you!

Yes. Here is a generalization:

Definition. Let $K$ be a field, and let $m$ and $n$ be two nonnegative integers. Let $A_1, A_2, \ldots, A_n$ be $n$ arbitrary $m\times m$-matrices over $K$. Let $\sigma$ be a permutation of $\left\{1,2,\ldots,n\right\}$. Write $\sigma$ as a product of disjoint cycles (including $1$-cycles!) as follows:
  \begin{align}
\sigma = \prod\limits_{j=1}^g \underbrace{\left(k_{j,1}, k_{j,2}, \ldots, k_{j,l_k}\right)}_{\text{an $l_k$-cycle}}
\end{align}
  (so that $\sigma\left(k_{j,1}\right) = k_{j,2}$ and $\sigma\left(k_{j,2}\right) = k_{j,3}$ and so on ... and $\sigma\left(k_{j,l_k}\right) = k_{j,1}$, and each integer between $1$ and $n$ appears exactly once in the total list of all $k_{j,i}$'s). 
  Now, set
  \begin{align}
\operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right)
= \prod\limits_{j=1}^g \operatorname{Tr}\left(A_{k_{j,1}} A_{k_{j,2}} \cdots A_{k_{j,l_k}} \right) .
\end{align}

For example, if $n = 6$, and if the permutation $\sigma \in S_6$ is written in cycle notation (including $1$-cycles) as $\sigma = \left(1, 3, 6\right) \left(2\right) \left(4, 5\right)$, then
\begin{align}
\operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_6\right)
= \operatorname{Tr}\left(A_1 A_3 A_6\right) \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_4 A_5\right) .
\end{align}
Make sure you understand why this definition of $\operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right)$ does not depend on the precise way how the cycles $\left(k_{j,1}, k_{j,2}, \ldots, k_{j,l_k}\right)$ have been written. (That is: If we rewrite an $l$-cycle $\left(p_1, p_2, \ldots, p_l\right)$ as $\left(p_{u+1}, p_{u+2}, \ldots, p_l, p_1, p_2, \ldots, p_u\right)$, then $\operatorname{Tr}\left(A_{p_1} A_{p_2} \cdots A_{p_l}\right) = \operatorname{Tr}\left(A_{p_{u+1}} A_{p_{u+2}} \cdots A_{p_l} A_{p_1} A_{p_2} \cdots A_{p_u}\right)$.)

Theorem 1. Let $K$ be a field, and let $m$ and $n$ be two nonnegative integers such that $n > m$. Let $S_n$ denote the group of all permutations of $\left\{1,2,\ldots,n\right\}$. For each permutation $\sigma \in S_n$, let $\left(-1\right)^\sigma$ be the sign of $\sigma$. Then,
  \begin{align}
\sum_{\sigma \in S_n} \left(-1\right)^\sigma \operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right) = 0 .
\end{align}

Your claim is the particular case of Theorem 1 for $n = 3$ and $m = 2$. In fact, the group $S_3$ of all permutations of $\left\{1,2,3\right\}$ has $6$ elements, which are
\begin{align}
\left(1\right)\left(2\right)\left(3\right), \qquad \left(1,2,3\right), \qquad \left(1,3,2\right), \\
\left(1\right)\left(2,3\right), \qquad \left(2\right)\left(1,3\right), \qquad \left(3\right)\left(1,2\right)
\end{align}
when written in the cycle notation; and their corresponding $\operatorname{Tr}_\sigma\left(A_1, A_2, A_3\right)$ are
\begin{align}
\operatorname{Tr}_{\left(1\right)\left(2\right)\left(3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_3\right) ; \\
\operatorname{Tr}_{\left(1,2,3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1 A_2 A_3\right) ; \\
\operatorname{Tr}_{\left(1,3,2\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1 A_3 A_2\right) ; \\
\operatorname{Tr}_{\left(1\right)\left(2,3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2 A_3\right) ; \\
\operatorname{Tr}_{\left(2\right)\left(1,3\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_1 A_3\right) ; \\
\operatorname{Tr}_{\left(3\right)\left(1,2\right)} \left(A_1, A_2, A_3\right) &= \operatorname{Tr}\left(A_3\right) \operatorname{Tr}\left(A_1 A_2\right) ; \\
\end{align}
and thus the claim of Theorem 1 rewrites as
\begin{align}
& \operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_3\right)
+ \operatorname{Tr}\left(A_1 A_2 A_3\right) 
+ \operatorname{Tr}\left(A_1 A_3 A_2\right) \\
&=
\operatorname{Tr}\left(A_1\right) \operatorname{Tr}\left(A_2 A_3\right) 
+ \operatorname{Tr}\left(A_2\right) \operatorname{Tr}\left(A_1 A_3\right) 
+ \operatorname{Tr}\left(A_3\right) \operatorname{Tr}\left(A_1 A_2\right)
\end{align}
in this case (for any three $2 \times 2$-matrices $A_1, A_2, A_3$). The latter equality is precisely your claim.
Theorem 1 can be proven using a bit of multilinear algebra. The rough idea is this: Let $V$ be the $K$-vector space $K^m$; thus each matrix $A_i$ becomes an endomorphism of $V$. Now, consider the endomorphism $A_1 \otimes A_2 \otimes \cdots \otimes A_n$ of the tensor power $V^{\otimes n}$. But permutations $\sigma \in S_n$ also act on $V^{\otimes n}$ by permuting the tensor factors:
\begin{align}
\sigma \left(v_1 \otimes v_2 \otimes \cdots \otimes v_n\right)
= v_{\sigma^{-1}\left(1\right)} \otimes v_{\sigma^{-1}\left(2\right)} \otimes \cdots \otimes v_{\sigma^{-1}\left(n\right)}
\end{align}
for all $v_1, v_2, \ldots, v_n \in V$ and $\sigma \in S_n$.
Now, you can show that
\begin{align}
\operatorname{Tr}_\sigma\left(A_1, A_2, \ldots, A_n\right)
= \operatorname{Tr} \left( \sigma^{-1} \circ \left(A_1 \otimes A_2 \otimes \cdots \otimes A_n\right) \right)
\end{align}
for any $\sigma \in S_n$, where the $\sigma^{-1} \circ \left(A_1 \otimes A_2 \otimes \cdots \otimes A_n\right) $ on the right hand side is understood to be an endomorphism of $V^{\otimes n}$. Hence, the claim of Theorem 1 will easily follow if we can show that
\begin{align}
\sum_{\sigma \in S_n} \left(-1\right)^\sigma \sigma^{-1} \circ \left(A_1 \otimes A_2 \otimes \cdots \otimes A_n\right) = 0 .
\end{align}
But this, in turn, follows from
\begin{align}
\sum_{\sigma \in S_n} \left(-1\right)^\sigma \sigma^{-1} = 0 \qquad \text{as endomorphisms of } V^{\otimes n}
\end{align}
(which I leave to you to check).
The theory behind this is known as "Schur-Weyl duality for the general linear group", but nothing deep is being used for this specific argument.
