 Solution

Writing $X=u$, $Y=v$, and $Z=-w$ turns the cubic into $X^3+Y^3=Z^3$. The birational coordinatesgive the Weierstrass equationThe original projective cubic has no common zero of its three partial derivatives modulo any $p\ne 3$, so it is already a smooth proper model and $C$ has good reduction at every such prime.

If $p\equiv 1(\mathrm{mod}3)$, then ${\mu }_3\subset {\mathbb{F}}_p$, all nine inflection points are rational, and $C({\mathbb{F}}_p)$ contains $(\mathbb{Z}/3\mathbb{Z}{)}^2$, so it is not cyclic. If $p\equiv 2(\mathrm{mod}3)$, cubing is a bijection on ${\mathbb{F}}_p$. Counting on the Fermat model gives $\#C({\mathbb{F}}_p)=p+1$. A finite elliptic-curve group has the form $\mathbb{Z}/d\mathbb{Z}\times \mathbb{Z}/e\mathbb{Z}$ with $d\mid e$ and $d\mid p-1$, so $d\mid \gcd (p-1,p+1)=2$. For odd $p$, the equation $x^3=432$ has exactly one root because cubing is bijective, so there is only one nonzero rational 2-torsion point and $d\ne 2$. Hence the group is cyclic. For $p=2$ it has order three and is cyclic as well.

Solved by gpt-5.6-sol high.