General expression of time-ordered thermal average Green's function does not reproduce non-interacting limit (Fetter ch. 31 Eq. (31.24))

Hi I am going through Fetter’s Quantum Theory of Many-Particle Systems Dover Edition.

In ch. 31 he computed the relation between $bar{G}(mathbf{k},omega)$, ${bar{G}}^{R}(mathbf{k},omega)$ and $bar{G}^{A}(mathbf{k},omega)$ where each of them are the real-time Fourier transformation of the corresponding real-time thermal averaged Green’s function. The thermal average is done in the grand canonical ensemble.

It is proved that we can express $bar{G}(mathbf{k},omega)$ for Fermi system as
begin{equation}
bar{G}(mathbf{k},omega) = [1 + e^{-beta hbar omega}]^{-1}{bar{G}}^{R}(mathbf{k},omega)+[1+e^{beta hbar omega}]^{-1} {bar{G}}^{A}(mathbf{k},omega).
end{equation}
This is in Eq. (31.24) of Fetter and I totally agree with this expression as I can really derive it. However, things get weird when I try to apply this expression to non-interacting Fermi system.

Suppose we have a non-interacting Fermi system with grand canonical Hamiltonian $K = sum_{mathbf{k}} left( epsilon_{mathbf{k}} – muright)c^{dagger}_{mathbf{k}}c_{mathbf{k}} = sum_{mathbf{k}}xi_{mathbf{k}}c^{dagger}_{mathbf{k}}c_{mathbf{k}}$ where $epsilon_{mathbf{k}}$ is the kinetic energy and $mu$ is the chemical potential, then it is also known that
begin{equation}
bar{G}(mathbf{k},omega) = frac{1}{1+e^{-beta xi_{mathbf{k}}}}cdotfrac{1}{omega – hbar^{-1}xi_{mathbf{k}}+ieta} + frac{1}{1+e^{beta xi_{mathbf{k}}}}cdotfrac{1}{omega – hbar^{-1}xi_{mathbf{k}} – ieta}, text{ (Eq. (31.38) in Fetter)}
end{equation}
begin{equation}
bar{G}^{R}(mathbf{k},omega) = frac{1}{omega – hbar^{-1}xi_{mathbf{k}}+ieta},
end{equation}
begin{equation}
bar{G}^{A}(mathbf{k},omega) = frac{1}{omega – hbar^{-1}xi_{mathbf{k}}-ieta},
end{equation}
where we will take $eta to 0^{+}$ in the end. However, it is not obvious to me that
begin{equation}
bar{G}(mathbf{k},omega) = [1 + e^{-beta hbar omega}]^{-1}{bar{G}}^{R}(mathbf{k},omega)+[1+e^{beta hbar omega}]^{-1} {bar{G}}^{A}(mathbf{k},omega).
end{equation}
can help us obtain the correct $bar{G}(mathbf{k},omega)$. This is because if we just plug in the non-interacting $bar{G}^{A}(mathbf{k},omega)$ and $bar{G}^{R}(mathbf{k},omega)$ which I have shown above, we will obtain
begin{equation}
bar{G}(mathbf{k},omega) = frac{1}{1 + e^{-beta hbar omega}}cdotfrac{1}{omega – hbar^{-1}xi_{mathbf{k}}+ieta}+frac{1}{1+e^{beta hbar omega}}cdot frac{1}{omega – hbar^{-1}xi_{mathbf{k}}-ieta}
end{equation}
which is not (or at least that that obvious?) equal to
begin{equation}
bar{G}(mathbf{k},omega) = frac{1}{1+e^{-beta xi_{mathbf{k}}}}cdotfrac{1}{omega – hbar^{-1}xi_{mathbf{k}}+ieta} + frac{1}{1+e^{beta xi_{mathbf{k}}}}cdotfrac{1}{omega – hbar^{-1}xi_{mathbf{k}} – ieta} text{ (Eq. (31.38) in Fetter)}.
end{equation}
In the first expression we still have a $omega$ dependence in $frac{1}{1 + e^{-beta hbar omega}}$ and $frac{1}{1+e^{beta hbar omega}}$ but in the second expression we have instead $frac{1}{1+e^{-beta xi_{mathbf{k}}}}$ and $frac{1}{1+e^{beta xi_{mathbf{k}}}}$. Are there suggestion to reconcile this? Thanks!