First step: When $p\equiv 1 \pmod{ 3}$, prove that there exists a pair $(a,b)$ of integers such that $4p=a^2+27b^2$, $a\equiv 1 \pmod{ 3}$ and a is unique (the proof of the first step).
Second step: Prove that $x^3+y^3=1$ has $p-2+a$ solutions in $\mathbb{F}_p$.
For the second step, we know that in $\mathbb{F}_p^*$, $\ \exists g\ $ s.t. $\mathbb{F}_p^*=\{g,g^2,\dots,g^{p-1}(=1)\}$. Hence $\{x^3|x\in\mathbb{F}_p^*\}=\{g^3,g^6,\dots,g^{p-1}\}$ is the unique subgroup of index 3 in $\mathbb{F}_p^*$. But I don't know the relation between the solution of the equation and the decomposition $4p=a^2+27b^2$.
Any ideas?