In 
 `http://131.220.77.52/files/preprints/diophantine/bruedern/wpminicubefinal.pdf`, Bruedern and Wooley mention the following fact on the bottom of page 6:
Let 
$$\Phi(\alpha) = \sum\_{h\le 6H}\sum\_{P<x\le 2P}e(\alpha h(3x^2 + 3xh + h^2))$$
They then assert that $$\int\_{0}^1\left|\Phi(\alpha)\right|^4d\alpha\ll H^3P^{2 + \epsilon}$$
for all $\epsilon>0$


However, I don't have access to the paper mentioned in which this fact is proven. Can someone show me a proof of this? It is supposed to be along the lines of the proof of Hua's Lemma, but I can't see how this is the case.



Exponential Sum Bound
First, the paper is Bruedern and Wooley!


Next ... by Cauchy's inequality, one has
$$|\Phi(\alpha)|^2\le 6H\Psi(\alpha),$$
say, where
$$\Psi(\alpha)=\sum\_{h\le 6H}\left| \sum\_{P<x\le 2P}e(\alpha h(3x^2+3xh+h^2))\right|^2.$$
Thus the mean value in question is bounded above by $6H$ times
$$\int\_0^1\Psi(\alpha)|\Phi(\alpha)|^2\, d\alpha ,$$
and by orthogonality that counts the number of solutions of the Diophantine equation
$$3h(x\_1-x\_2)(x\_1+x\_2+h)=g\_1(3y\_1^2+3y\_1g\_1+g\_1^2)-g\_2(3y\_2^2+3y\_2g\_2+g\_2^2),$$
with $h,g\_1,g\_2\le 6H$ and $P<x\_1,x\_2,y\_1,y\_2\le 2P$. When $x\_1=x\_2$ and $h$ is free, then given $g\_2$ and $y\_2$, one finds that $g\_1$ and $y\_1$ are determined by a divisor estimate. So the number of diagonal solutions is $\ll HP\cdot (HP)^{1+\epsilon}=(HP)^{2+\epsilon}$. Meanwhile, when $x\_1\ne x\_2$, the right hand side here is also non-zero. Then given a fixed choice for $g\_1,g\_2,y\_1,y\_2$, one finds that $x\_1-x\_2$ and $h$ are determined by a divisor function estimate. There are consequently $\ll (HP)^2\cdot (HP)^\epsilon$ such solutions. Combine these estimates, throw back in the factor $6H$ that arose from applying Cauchy's inequality, and one gets the estimate claimed.


