Absolute value of a cubic Gauss sum (over Field $\mathbb{F}_p$ ) I'm interested in the quantity $|\sum_{x \in \mathbb{F}_p} \omega^{ax^3+bx^2+cx}|$ where $a \in \mathbb{F}_p^*,b,c \in \mathbb{F}_p$ and $\omega$ is a primitive $p$-th root of unity i.e. $\omega=e^{\frac{2 \pi i}{p}}$.
I know that the case $a=0$ is well understood and evaluates to $\sqrt{p}$ (quadratic Gauss sum).
When $a\neq 0$ I have observed numerically that, for fixed $a$, the set
$$ \mathcal{S}_b:= \{|\sum_{x \in \mathbb{F}_p} \omega^{ax^3+bx^2+cx}|, c \in \mathbb{F}_p \} $$
obeys
$$ \mathcal{S}_b =  \mathcal{S}_{b^\prime\neq b}$$
In other words, it appears that the quadratic coefficient is somehow unimportant for my purposes.
Question: (i) Is there some way of manipulating my expression to see why this property holds? or, (ii) Is there a relevant reference that states something along these lines?
 A: Let us assume that $\Bbb{F}_p$ does not have characteristic three (so $p>3$ if you are, as seems to be the case, only interested in prime fields). Denote
$$
T(a,b,c)=\sum_{x\in\Bbb{F}_p}\omega^{ax^3+bx^2+cx}.
$$
Assume that $a\neq0$. Let $z\in\Bbb{F}_p$ be a parameter. For a fixed $z$ as $x$ ranges over $\Bbb{F}_p$ so does $x+z$. Thus
$$
\begin{aligned}
T(a,b,c)&=\sum_{x\in\Bbb{F}_p}\omega^{a(x+z)^3+b(x+z)^2+c(x+z)}\\
&=\sum_{x\in\Bbb{F}_p}\omega^{[ax^3+(b+3az)x^2+(c+2bz+3az^2)x]+[az^3+bz^2+cz]}\\
&=\omega^{az^3+bz^2+cz}T(a,b+3az,c+2bz+3az^2).
\end{aligned}
$$
In particular (any power of $\omega$ has absolute value $=1$)
$$
|T(a,b,c)|=|T(a,b+3az,c+2bz+3az^2)|.
$$
Here $3a$ is an invertible element of $\Bbb{F}_p$, so by varying $z$, the second parameter on the right hand side, $b+3az$, ranges over the entire field. The particular choice $z_0=-b/3a$ makes it zero giving us
$$
|T(a,b,c)|=|T(a,0,c+2bz_0+3az_0^2)|.
$$
Fixing $a,b$ (and thus also $z_0$) and varying $c$ shows then that
$$
S(b)=S(0),
$$
because $c$ ranges over the elements of $\Bbb{F}_p$ as $c+2bz_0+3az_0^2$ does.

This kind of linear substitution tricks are common in manipulating character sums over finite fields. A more general linear substitution would allow you to replace $a$ with any other element $a'$ such that $a'/a$ is a non-zero cube.
