Let $C(x_1,\ldots,x_n)$ be a nonsigular cubic form with integral coefficients.
In his Proof that $C$ fulfills the Hasse-Principle, if $n\geq 9$, Hooley used the following estimate that was provided by Katz in *Perversity and Exponential Sums*:

There exists a subset $\mathcal{P}$ of primes, having positive Dirichlet density and $A<1$ such that for all $w\in \mathcal{P}$ we have $$ \sum_{\mathbf{b}(\text{mod } w)} |\sum_{\substack{t (\text{mod } w)\\ (t,w)=1}} \sum_{\mathbf{x} (\text{mod } w)}e_w(tC(\mathbf{x})+\mathbf{b}.\mathbf{x})|<Aw^{(3n+1)/2} $$ where $\mathbf{b}$ is the vector $(b_1,\ldots b_n)$, $e_q(z)=e^{2 \pi \text{i} z/q}$ and $\mathbf{a}.\mathbf{b}=\sum_{i=1}^n a_i b_i$.

Question: Can we do the same when we twist the innermost sum by $\chi(t)$ for a multiplicative character $\chi$? (I need it only in the case where $\chi$ is the legendre symbol.)

Context: I try to prove the Hasse-Principle for $C(\mathbf{x})=y^2$ and $n=6$. Everything goes through nicely up to the aforementioned problem.

Looking at prime moduli and not averaging over $\mathbf{b}$, there were sufficient results (provided by Katz) that showed that the additional character does not matter, but I was not able to find it in the averaged case. My own Algebraic-Geometry knowledge is insufficient for trying to change the proof of *Perversity and Exponential sums*.