An exact counterexample to the arbitrary-exponent determinant extension

The determinant comparison $\det(A^k+|AB|^p)\ge\det(A^k+|BA|^p)$, for
positive definite $A$, invertible Hermitian $B$ and $k,p\ge0$, fails for real
matrices of order three with $k=6$ and $p=8$. One may take
$A=\operatorname{diag}(2,1,1/2)$ and
$B=\frac15\begin{pmatrix}-1&2&0\\2&1&2\\0&2&1\end{pmatrix}$, which is
symmetric with $\det B=-1/125$. Since $|AB|^2=BA^2B$ and $|BA|^2=AB^2A$, both
sides are determinants of $A^6+5^{-8}K$ with $K=(MA^2M)^4$ or $(AM^2A)^4$,
$M=5B$. Expanding $\det(D+zK)$ by multilinearity in dimension three gives an
exact two-coefficient polynomial for the difference, and substituting
$z=5^{-8}$ shows that the right side exceeds the left by
$21036678407451/156250000000000$. The example uses an indefinite $B$; it does
not contradict the known case $k=2$ and does not address the variant with
$B$ positive definite.