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Let $ S(\alpha)=\sum_{x\leq X}\sum_{y\leq Y}e \left(\alpha x y^3 \right)$, for some $X,Y \geq 1$ and write $\alpha = a/q + \beta$ for $(a,q)=1$, as usual.

For the 'classical' cubic Weyl-sum $f(\alpha)= \sum_{y\leq Y} e\left(\alpha y^3 \right)$ on has a bound due to Vaughan (The Hardy-Littlewood Method - Theorem 4.1) of the form $$ f(\alpha) \ll q^{-1/3} Y \left(1+Y^3 |\beta | \right)^{-1/3} + q^{1/2+\epsilon} \left(1+Y^3 |\beta| \right)^{1/2}, $$ where the first part comes from the usual major arc approximation as $$q^{-1}\sum_{x=1}^q e\left(\frac{ax^3}{q}\right) \int_{0}^Y e(\beta y^3) dy.$$

I am looking for a similar bound for the sum $S(\alpha)$, i.e. an approximation of the form $$ S(\alpha) = q^{-2} \sum_{x=1}^q \sum_{y=1}^q e\left(\frac{axy^3}{q} \right) \int_{1}^X \int_{1}^Y e(\beta xy^3) dx dy + O\left(q^{\theta+\epsilon} \left(1+XY^3 |\beta| \right)^{\theta} \right).$$

The standard approach via partial summation should give $\theta =2$, but for the application in mind - that is an $\epsilon$-free Weyl's inequality, I need preferably something like maybe $\theta = 2/3$. Is it even possible to go below $\theta=1$?

If I try to make an analogous version of Vaughan's proof involving Poisson summation work, there seem to be several problems going one dimension higher.

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