= Solution
<Mass> preservation of a nonnegative finite-<mass> profile requires $\alpha+\beta=0$. Balancing $\rho_t$ and $\partial_x(\rho^4/4)$ requires $\alpha-1=4\alpha-\beta$. Therefore
$$
\boxed{\alpha=-\tfrac14,\qquad\beta=\tfrac14.}
$$
The similarity ODE is
$$
-\tfrac14(f+\eta f')+f^3f'=0,\qquad (f^4-\eta f)'=0.
$$
For a source-type profile adjoining vacuum, the integration constant is zero. The positive branch is $f=\eta^{1/3}$ for positive $\eta$, terminated by a <shock> into zero. Normalize the <mass>:
$$
1=\int_0^{\eta_F}\eta^{1/3}\,d\eta=\frac34\eta_F^{4/3},\qquad \eta_F=(4/3)^{3/4}.
$$
Thus a unit-<mass> <similarity solution> is
$$
\boxed{\rho(x,t)=t^{-1/4}\begin{cases}(x/t^{1/4})^{1/3},&0<x<\eta_Ft^{1/4},\\0,&\text{otherwise}.\end{cases}}
$$
At the moving right edge, $S'=\eta_Ft^{-3/4}/4$ equals $\rho_L^3/4$, so the <Rankine-Hugoniot condition> is satisfied. The <characteristic speeds> satisfy $0<S'<\rho_L^3$, making it an entropy <shock>. The left edge meets vacuum continuously. As $t\downarrow0$ the <mass> concentrates at the origin; this is a source solution, not a claim that every finite-<mass> initial profile equals it at finite time.
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