= Solution
Away from the three poles, the identity from the preceding part puts $F(u)$ on the cubic $Q=0$. The local extensions and continuity put the three pole images there too. Thus \b[$F(C)\subseteq\{Q=0\}$], a cubic curve.
The local calculations gave
$$
F(p)=[0:1:0],\qquad F(q)=[0:b:c],\qquad F(r)=[0:0:1],\quad bc\ne0.
$$
They are three distinct points on the line with first coordinate zero. Hence \b[$F(p),F(q),F(r)$ are collinear]. This is <three-point Riemann-Roch construction of a plane cubic>.
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