= Solution
Write $H=\dot a/a$ for the <Hubble parameter>. Subtracting the <Friedmann equation> from the second equation gives the <Friedmann acceleration equation>,
$$
\frac{\ddot a}{a}=-\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)+\frac{\Lambda c^2}{3}.
$$
Since $\dot H=\ddot a/a-H^2$, eliminating $H^2$ with the first equation yields
$$
\dot H-\frac{kc^2}{a^2}=-4\pi G\left(\rho+\frac p{c^2}\right).
$$
Differentiating the first equation, with constant <cosmological constant> and curvature parameter, now gives
$$
\frac{8\pi G}{3}\dot\rho=2H\left(\dot H-\frac{kc^2}{a^2}\right)=-8\pi GH\left(\rho+\frac p{c^2}\right).
$$
Thus the <cosmological perfect-fluid continuity equation> is $\dot\rho+3H(\rho+p/c^2)=0$. Multiplying by $a^3$ and using $d(a^3)/dt=3Ha^3$ proves
$$
\boxed{\frac{d}{dt}(\rho a^3)+\frac p{c^2}\frac{d}{dt}(a^3)=0.}
$$
This is local <conservation of energy>: expansion changes the energy in a <comoving volume> through pressure work.
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