Let $p$ be an odd prime and $F(x,y)$ be a binary cubic form such that $F(x,y) \equiv a(x-by)(x-cy)(x-dy) \hspace{1mm} (\text{mod } p)$. Then the curve $z^{3} = F(x,y)$ has a non-zero solution in the $\mathbb{Z}_{p}$, the space of $p$-adic numbers.
When $p \equiv 2 \hspace{1mm} (\text{mod } 3)$, the result follows because every residue class is a cubic-residue. Then I can lift the solution using Hensel's lemma. I am struggling to see it when $p \equiv 1 \hspace{1mm} (\text{mod} 3)$.
I know of a slightly high-level proof :The curve $z^{3} = F(x,y)$ is smooth over primes $p \equiv 1 \hspace{1mm} (\text{mod } 3)$ and has genus $\binom{3-1}{2} = 1$. Therefore, using the Hasse-bound I could surmise that if $N$ is the number of points on the projective curve then
$$-2\sqrt{p} \leq N - (p+1) \leq 2\sqrt{p} $$ which gives that $$1 \le p - 2\sqrt{p} + 1 \leq N \leq p + 2\sqrt{p} + 1.$$
I am interested in knowing an elementary proof, as compared to what I have put up for $p \equiv 1 \hspace{1mm} (\text{mod } 3)$. Also what can I do for $p = 3$?