It is well known that, for symmetric matrices, the trace the product of three matrices $A$, $B$, $C$ is the same, independent of the order of $A$, $B$, $C$, that is,
$$\operatorname{tr}(ABC) = \operatorname{tr}(ACB) = \operatorname{tr}(BAC) = \operatorname{tr}(BCA) = \operatorname{tr}(CAB) = \operatorname{tr}(CBA)$$
And, that the trace is invariant under circular shifts,
$$\operatorname{tr}(ABCD) = \operatorname{tr}(DABC)$$
but, in general, it is not possible to make any permutation of we consider more than three factors.
I am studying the power of Laplacian matrix $L^m =(D-A)^m$, where $D$ is the degree matrix of a graph and $A$ its adjacency matrix. When opening $(D-A)^m$, I want to relate all factors that has same number of $D$’s and $A$’s together, like when opening $(d-a)^m$ for $a,d\in\mathbb{R}$, which results in $$\sum_{i=0}^m{m\choose j}(-1)^jd^{m-j}a^j.$$
For example, $m=5$, is there any relation of $\operatorname{tr}(ADADA)$ and $\operatorname{tr}(D^2A^3)$? Something $\operatorname{tr}(ADADA) = \operatorname{tr}(D^2A^3) + o(n^5)$.
Any help is appreciated!