This is a conjecture, refined through previous responses, about sums of positive integers which define sums of the corresponding cubes.

Question

Revising $4255316$, the conjecture now is: If $$\sum_{i=1}^{i=n}{a_{i}}, i>2$$, divides $$3\sum{x_{i}x_{j}x_{k}}$$, where the product is summed over the$\binom{n}{3}$distinct triplets drawn from the $n$ suffices, then $$\sum_{y=0}^{y=a}\sum_{i=1}^{i=n}{x_{i}+2yb}$$ divides $$\sum_{y=0}^{y=a}\sum_{i=1}^{i=n}{(x_{i}+2yb)^{3}}$$