Simplify exponential sum over $\mathbb{F}_p$ to prove identity I have a sum involving  $p$-th roots of unity (where $\frac{1}{t}$ is to be understood as the field inverse $t^{-1} \bmod p$ etc.) of the form
$\begin{align*}
&d_{j,k}=\sum_{a,b,c \in \mathbb{F}_p}\omega^{\frac{1}{12}(a^3-c^3)+(\frac{b}{2}+\frac{1}{8})(a^2-c^2)+(\frac{b}{2}+b^2)(a-c)+kc-ja}\end{align*}$
where $\omega=\exp(2 \pi i/p)$ is a primitive $p$-th root of unity and $p$ is a prime greater than 3.
With the aid of a computer, I can confirm the above expression obeys
$\begin{align*}d_{j,k\neq j}=p^2 \sum_{r  \in \mathbb{F}_p}\omega^r=0
\end{align*}$
I have checked this for $5\leq p \leq 17$ and am confident this holds for all prime $p>3$.
Question: 
How can I manipulate the expression for $d_{j,k}$ in order to see that all $d_{j,k\neq j}$ (with $j,k \in \mathbb{F}_p$) evaluate to zero? 
 A: The character sum involves the polynomial
$$
f(a,b,c)=\frac{1}{12}(a^3-c^3)+(\frac{b}{2}+\frac{1}{8})(a^2-c^2)+(\frac{b}{2}+b^2)(a-c)+kc-ja.
$$
Staring at this for a while suggested to me to look at
$$
f(A+C,B-A/2-1/4,A-C)=\frac{C^3}6+C\left(-\frac18+2B^2-j-k\right)+A(-j+k)
$$
instead (ain't Mathematica's Simplify wonderful!).
In the interest of avoiding those tiny double supercripts let me define $e(x)=\omega^x$ for all $x\in\Bbb{F}_p$.
The linear substitution $(A,B,C)\mapsto (A+C,B-A/2-1/4,A-C)$ is bijective because $p>2$. Therefore we might as well use $(A,B,C)$ as summation variables and let them range over the entire field. We get
$$
\begin{aligned}
d_{j,k}&=\sum_{A,B,C\in\Bbb{F}_p}e\left(\frac{C^3}6+C\left(-\frac18+2B^2-j-k\right)+A(-j+k)\right)\\
&=\sum_{B,C\in\Bbb{F}_p}e\left(\frac{C^3}6+C\left(-\frac18+2B^2-j-k\right)\right)
\sum_{A\in\Bbb{F}_p}e(A[k-j]).
\end{aligned}
$$
Here I used the fact that $e(x+y)=e(x)e(y)$, which allowed us to take that $A$-sum as a common factor.
Anyway, if $k\neq j$, then the $A$-sum is trivially zero. Thus $d_{j,k}=0$ unless $k=j$.
The key to success was the ability to separate that simple $A$-sum.
