= Solution
Assume fixed mass, constant $\Lambda$, no energy loss or surface work, and the same uniform-sphere structure coefficient at turnaround and virialization. Set $x=r_v/r_m$ and define the density in $\eta$ to be the mean turnaround density,
$$
\rho_m=\frac{3M}{4\pi r_m^3},\qquad
\eta=\frac{\Lambda}{4\pi G\rho_m}=\frac{\Lambda r_m^3}{3GM}.
$$
It cannot be the background or final density if the displayed cubic is to hold with these coefficients. Put $A=3GM^2/(5r_m)>0$ and $B=\Lambda Mr_m^2/10=(\eta/2)A$. The turnaround energy is $E_m=-A-B$. At virialization, the relation in part (c) gives
$$
E_v=K_v+U_{G,v}+U_{\Lambda,v}
=\frac12U_{G,v}+2U_{\Lambda,v}
=-\frac{A}{2x}-2Bx^2.
$$
Equating the two energies and dividing by $A$ yields $1+\eta/2=1/(2x)+\eta x^2$. Multiplication by $2x$ gives
$$
\boxed{2\eta x^3-(2+\eta)x+1=0,\qquad x=r_v/r_m.}
$$
This derivation also fixes the density convention that the printed statement leaves implicit.
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