Let $\mathbb{F}_q$ be a finite field of $q$ elements. Let $A,B,C,\cdots$ denote the multiplicative character 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][1], 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 Quesition 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!

  [1]: https://doi.org/10.1090/S0002-9947-1987-0879564-8