= Cubic eighth-moment proof by differencing
{title2=$\int_0^1|\sum_{x\le n}e(\theta x^3)|^8\,d\theta\ll_\eta n^{5+\eta}$}
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 $D\ne0$, 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 $|S|^4\le2n\sum_v c_ve(\theta v)$, 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 $v\ne0$: 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 $\sum_vr(v)=n^4$. Multiply the inequality by $|S|^4$ and integrate. It follows that $I_8\ll n(n^2I_4+n^\eta n^4)\ll_\eta n^{5+\eta}$, after choosing the divisor-bound exponents small enough.
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