You are asking for a polynomial equality in 16 variables. A polynomial equality holds if it holds on an open subset of affine 16-space. So you can restrict to the open subset of 16-space where A, B, C and D have distinct eigenvalues. Now because they commute they can be simultaneously diagonalized (perhaps over a bigger field, but that's okay). The truth of your equation is not affected if we apply this diagonalization (you are conjugating each matrix by a single matrix $X$, which does not change either side of the equation). So you can assume A, B, C and D are all diagonal, which makes this an easy computation.
You are asking for a polynomial equality in 16 variables. A polynomial equality holds if it holds on an open subset. So you can restrict to the open subset of 16-space where A, B, C and D have distinct eigenvalues. Now because they commute they can be simultaneously diagonalized (perhaps over a bigger field, but that's okay). The truth of your equation is not affected if we apply this diagonalization (you are conjugating each matrix by a single matrix $X$, which does not change either side of the equation). So you can assume A, B, C and D are all diagonal, which makes this an easy computation.
You are asking for a polynomial equality in 16 variables. A polynomial equality holds if it holds on an open subset of affine 16-space. So you can restrict to the open subset of 16-space where A, B, C and D have distinct eigenvalues. Now because they commute they can be simultaneously diagonalized (perhaps over a bigger field, but that's okay). The truth of your equation is not affected if we apply this diagonalization (you are conjugating each matrix by a single matrix $X$, which does not change either side of the equation). So you can assume A, B, C and D are all diagonal, which makes this an easy computation.
You are asking for a polynomial equality in 16 variables. A polynomial equality holds if it holds on an open subset. So you can restrict to the open subset of 16-space where A, B, C and D have distinct eigenvalues. Now because they commute they can be simultaneously diagonalized (perhaps over a bigger field, but that's okay). The truth of your equation is not affected if we apply this diagonalization (you are conjugating each matrix by a single matrix $X$, which does not change either side of the equation). So you can assume A, B, C and D are all diagonal, which makes this an easy computation.