Why $\rho(AABABB)=\rho(ABAABB)$? Let $A,B$ be two matrices, $\rho$ be spectral radius, which is the top eigenvalue of a matrix. I discovered that
$$\rho(AABABB)=\rho(ABAABB).$$
But I could not find the reason.
By the way, all I had tested were $2\times2$ positive matrices. And I know the fact $\rho(AB)=\rho(BA)$, but it seems that this property is not enough for this statement.
 A: Your statement is true for all $2\times2$ matrices (positive or not). Recall that matrix trace of matrix product is invariant under cyclic permutation. Let $\renewcommand{\tr}{\operatorname{tr}}t=\tr(AB)=\tr(BA)$ and $d=\det(AB)=\det(BA)$. Then
\begin{align}
\renewcommand{\tr}{\operatorname{tr}}
\tr(AABABB) &= \tr[(AB)^2 BA] = \tr[(tAB-dI)BA] = \tr(tAABB - dAB),\\
\tr(ABAABB) &= \tr[(BA)^2 AB] = \tr[(tBA-dI)AB] = \tr(tAABB - dAB).
\end{align}
Hence the two matrix products have identical traces. Obviously they have identical determinants too. So, by Cayley-Hamilton theorem, they have identical spectra and in turn identical spectral radii.
I guess the statement is not true for larger-sized matrices, but I don't have time to perform any computational test.
A: Here is a less satisfactory negative answer using Octave:
octave:1> a=rand(3,3)
a =

   0.853184   0.968858   0.369978
   0.995402   0.194116   0.449798
   0.373102   0.045246   0.894742

octave:2> b=rand(3,3)
b =

   0.60655   0.11672   0.90867
   0.30900   0.89411   0.17607
   0.14202   0.94938   0.92741

octave:3> abs(eig(a*a*b*a*b*b))
ans =

   23.223839
    0.035014
    0.034103

octave:4> abs(eig(a*b*a*a*b*b))
ans =

   22.749277
    0.032287
    0.037755

