= Solution
The <photon number density> is obtained by integrating the <Planck photon distribution>. With $x=h\nu/(kT)$,
$$
n=\frac{8\pi}{c^3}\int_0^\infty\frac{\nu^2\,d\nu}{e^{h\nu/(kT)}-1}
=\underbrace{\frac{8\pi k^3}{c^3h^3}\int_0^\infty\frac{x^2}{e^x-1}\,dx}_{\alpha}\,T^3.
$$
The integral converges, since its integrand is of order $x$ near zero and decays exponentially at infinity. Hence \b[$n=\alpha T^3$], with $\alpha$ independent of temperature.
For equilibrium radiation undergoing <adiabatic expansion>, the total <entropy> in a comoving volume is constant. Since the <entropy density> is $\beta T^3$, this means $\beta T^3V$ is constant. A comoving volume scales as the cube of the <scale factor>, so
$$
\boxed{T\propto V^{-1/3}\propto a^{-1}}.
$$
This agrees with the <cosmological redshift> of each photon frequency. The conclusion uses entropy conservation during the expansion.
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