= Solution
For a reversible change of a simple compressible system with fixed composition, the first law is
$$
\boxed{dE=T\,dS-P\,dV.}
$$
An adiabatic change exchanges no heat. A reversible adiabatic change is isentropic, $dS=0$; adiabatic alone does not imply constant <entropy> for an irreversible process.
Define the <Helmholtz free energy> $F=E-TS$. Then $dF=-S\,dT-P\,dV$. Equality of its mixed partial derivatives gives the <Maxwell relation>
$$
\boxed{\left(\frac{\partial S}{\partial V}\right)_T
=\left(\frac{\partial P}{\partial T}\right)_V.}
$$
Holding $T$ fixed in the first law and using this relation gives
$$
\boxed{\left(\frac{\partial E}{\partial V}\right)_T
=T\left(\frac{\partial P}{\partial T}\right)_V-P.}
$$
For equilibrium radiation, $E=Ve(T)$ and $P=e(T)/3$, so this identity reads $e=Te'/3-e/3$, or $Te'=4e$. Integrating gives
$$
\boxed{e=aT^4,\qquad E=aVT^4.}
$$
Finally in a reversible adiabatic expansion, $dE=-P\,dV$ becomes $4aVT^3dT+aT^4dV=-aT^4dV/3$. Dividing gives $3\,dT/T=-dV/V$, hence \b[$VT^3$ is constant]. This assumes equilibrium radiation throughout the expansion.
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