 Solution

Expand the secular Poynting–Robertson drag rates using $(1-e^2{)}^{-3/2}=1+\frac{3}{2}{e}^2+O(e^4)$ and $(1-e^2{)}^{-1/2}=1+O(e^2)$. Through second order in orbital eccentricity,The second formula follows from $d(e^2)/dt=2e\dot{e}$. The cubic correction in $\dot{e}$ contributes only at fourth order in $d(e^2)/dt$.