Base case: Computing the factorization requires $\frac 2 3n^3$ operations and the inverse requires $\frac 4 3n^3$ operations. Computing the trace adds $O(n)$. Let's round that to $2n^3$.
LU case: Computing the factorization requires $\frac 2 3n^3$ operations. If you were to compute $L^{-1}$ and $U^{-1}$, those operations each require $\frac 1 3n^3$ operations. The trace of the matrix product $U^{-1}L^{-1}$ is essentially a dot product of two vectors with length $n^2$, call that $O(n^2)$. Let's round that to $\frac 4 3n^3$.
Asymptotically, you're no better off. But, you can calculate the trace with approximately 33% fewer operations.
Source: https://software.intel.com/content/www/us/en/develop/documentation/mkl-developer-reference-fortran/top/lapack-routines/lapack-linear-equation-routines/lapack-linear-equation-computational-routines.html