Let $p$ be an odd prime, $\chi$ a Dirichlet character of order 3 over $\mathbb{F}_p-\{0\}$ and $\rho$ the quadratic character modulo $p$.
Show that $g(\chi)^3=p\chi(2)J(\chi,\rho)$, where $g$ is the Gauss Sum and $J$ is the Jacobi Sum.
I've already proved that $g(\chi)^2=\bar\chi(4)J(\chi,\rho)g(\chi^2)$ and I tried the following:
$g(\chi)^3=\bar\chi(4)J(\chi,\rho)g(\chi^2)g(\chi)=\bar\chi(4)J(\chi,\rho)g(\chi^{-1})g(\chi)$
but I don't know what to do next, cause if I substitute $g(\chi^{-1})g(\chi)$ by $J(\chi^{-1},\chi)g(\chi^{-1}\chi)$ then everything goes to cero because $g(\chi^{-1}\chi)=g(\epsilon)=0$.
Any suggestion?