 Cubic eighth-moment proof by differencing

Write $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $I_j={\int }_0^1|S{|}^j$. Character orthogonality gives $I_2=n$. For $I_4$, fix the difference $D=x^3-y^3$. If $D=0$, there are $n^2$ choices for the two equal pairs. If $D\ne 0$, any solution $z^3-w^3=D$ has $z-w\mid D$; fixing this divisor gives a quadratic equation for $w$, hence at most two choices. The subpower bound for the divisor function yields $I_4{\ll }_{\eta }{n}^{2+\eta }$.

The twice-differenced inequality in Weyl differencing reads $|S{|}^4\le 2n{\sum }_v{c}_{v}e(\theta v)$, with $c_v$ counting triples $(h,l,x)$ for which $v=3hl(2x+h+l)$ and all four shifted arguments are in $[1,n]$. Thus $c_0\ll n^2$, and $c_v{\ll }_{\eta }{n}^{\eta }$ for $v\ne 0$: the product $hl$ divides $v$, and the remaining equation determines $x$. The Fourier coefficients $r(v)$ of $|S{|}^4$ are nonnegative counts, with $r(0)=I_4$ and ${\sum }_{v}r(v)=n^4$. Multiply the inequality by $|S{|}^4$ and integrate. It follows that $I_8\ll n(n^2{I}_4+n^{\eta }{n}^4){\ll }_{\eta }{n}^{5+\eta }$, after choosing the divisor-bound exponents small enough.