Undergrad Separability of a Hamiltonian with spin

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The Hamiltonian in question, expressed as ##\hat{H}=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2+\frac{A}{\hbar^2}(J^2-L^2-S^2)##, can be separated into two parts: ##H_1## and ##H_2##. The commutator ##[H_1,H_2]## evaluates to zero, indicating that the two parts commute. The calculations show that both the momentum squared and the position squared operators commute with angular momentum and spin operators. Thus, the Hamiltonian is indeed separable into ##H_1 + H_2##. This confirms the separability of the Hamiltonian with respect to the defined components.
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I'd like to know if this Hamiltonian ##\hat{H}=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2+\frac{A}{\hbar^2}(J^2-L^2-S^2)## is separable into two parts ##H_1=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2## and ##H_2=\frac{A}{\hbar^2}(J^2-L^2-S^2)## and ##[H_1,H_2]=0##. Here A is a constant. I did so:

##[H_1,H_2]=[\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2,\frac{A}{\hbar^2}(J^2-L^2-S^2)]=##
##=[\frac{p^2}{2m},\frac{A}{\hbar^2}(J^2-L^2-S^2)]+[\frac{1}{2}m\omega^2r^2,\frac{A}{\hbar^2}(J^2-L^2-S^2)]=##
##=\frac{A}{2m \hbar^2}([p^2,J^2]-[p^2,L^2]-[p^2,S^2])+\frac{a}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])=## ##=\frac{A}{2m \hbar^2}([p^2,L^2]+[p^2,S^2]+2[p^2,L_xS_x]+2[p^2,L_yS_y]+2[p^2,L_zS_z]-[p^2,L_x^2]-[p^2,L_y^2]-[p^2,L_z^2]-## ##-[p^2,S^2])+\frac{A}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])=\frac{A}{2m \hbar^2}([p^2,2L_xS_x]+[p^2,2L_yS_y]+[p^2,2L_zS_z])+\frac{A}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])## now, since ##p^2## is a scalar and it's a function of spatial coordinates, it commutes with the component ##L_i## of the angular moment and with Spin operators and the same can be said about ##r^2## so the first commutator is ##0## and the hamiltonian is separable in ##H_1+H_2##. Am I right?
 
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