Let $\mathbb{F}_q$ be a finite field of $q$ elements. Let $A,B,C,\cdots$ denote the multiplicative characters over $\mathbb{F}_q$, and let $\overline{A}$ denote the inverse of $A$, i.e., $A(x)\overline{A}(x)=1$ for any $x\ne0$. Moreover for any multiplicative character $A$ we also define $A(0)=0$. In Greene's paper Hypergeometric functions over finite fields, for any multiplicative characters $A,B$, he defined $\binom{A}{B}$ by $$\binom{A}{B}=\frac{B(-1)}{q}J(A,\overline{B}),$$ where $J(A,\overline{B})$ is the Jacobi sum with respect to $A$ and $\overline{B}$.
In Definition 3.5 of Greene's paper, he defined the hypergeometric function $$_{2}F_1\left(\begin{align*}A,&B\\&C\end{align*}\mid x\right):=\epsilon(x)\sum_{y\in\mathbb{F}_q}B(y)\overline{B}C(1-y)\overline{A}(1-xy),$$ where $\epsilon$ is the trivial character. Also in Theorem 3.6 of the same paper, he showed that $$_{2}F_1\left(\begin{align*}A,&B\\&C\end{align*}\mid x\right)=\frac{q}{q-1}\sum_{\chi}\binom{A\chi}{\chi}\binom{B\chi}{C\chi}\chi(x).$$
In (4.11) of his paper, he proved that $$\binom{C}{A}+\binom{\varphi C}{A}=_{2}F_1\left(\begin{align*}A,\ &C^2\\&\overline{A}C^2\end{align*}\mid -1\right),$$ where $\varphi$ is the unique character of order $2$.
My Question is: Suppose $q\equiv1\pmod3$ and let $\psi$ be a character of order $3$. Motivated by the above, can we obtain an explicit value of $$\binom{C}{A}+\binom{\psi C}{A}+\binom{\psi^2 C}{A}?$$
Your comments are welcome!
