Well, I played myself with this formula and proved that if the following ratio holds:
$$\frac{a^3 + b^3 + c^3}{abc} = 6$$
then the three integers:
$$x_1 = a + b$$
$$x_2 = a + c$$
$$x_3 = b + c$$
are such that: $$ x_1^3 + x_2^3 + x_3^3 = y^3 $$
Proof is very simple:
Compute: $$ (a+b+c)^3$$
this gives: $$a^3 + b^3 + c^3 + 3[a^2(b+c) + b^2(a+c) + c^2(a+b)] + 6abc$$
now gather these addendums in three variables, forgetting about "6abc" for a while:
$$\alpha = a^3 + 3a^2b + 3ab^2$$
$$\beta = b^3 + 3b^2c + 3bc^2$$
$$\gamma = c^3 + 3c^2a + 3ca^2$$
now complete the gathering, adding to each variable a fraction of "6abc", obtaining:
$$x_1^3 = a^3 + 3a^2b + 3ab^2 + k_1abc$$
$$x_2^3 = b^3 + 3b^2c + 3bc^2 + k_2abc$$
$$x_3^3 = c^3 + 3c^2a + 3ca^2 + k_3abc$$
these cubes are perfect if and only if:
$$k_1abc = b^3$$
$$k_2abc = c^3$$
$$k_3abc = a^3$$
with the additional condition: $$k_1 + k_2 + k_3 = 6$$ where $$ k_1, k_2, k_3$$ are real
Now, the following equivalence holds:
$$a^3+b^3+c^3=6abc$$
and so:
$$\frac{a^3+b^3+c^3}{abc}=6$$
when this holds, the three numbers become:
$$x_1^3 = (a+b)^3$$
$$x_2^3 = (b+c)^3$$
$$x_3^3 = (c+a)^3$$
and then we have our sum of cubes.
Aftermaths!
This procedure does not say how to find the three numbers, but indeed once you find them, it is easy to show that each group of three base numbers, multiplied by an integer "h" still gives a sum of cubes that results in a perfect cube.
As an example, say:
(a, b, c) = (1, 2, 3)
then:
$$x_1 = 1 + 2 = 3$$
$$x_2 = 2 + 3 = 5$$
$$x_3 = 3 + 1 = 4$$
but also:
$$(a, b, c) = h(1, 2, 3)$$ with $$h \in \mathbb R$$ is a solution. Proof is starightforward:
if $$\frac{a^3+b^3+c^3}{abc}=6$$ holds, then also
$$\frac{h^3a^3+h^3b^3+h^3c^3}{hahbhc}=6$$ does. In fact you can group $$h^3$$ both above and beneath and then simplify.
One last thing to mention. Of course, while this does not help us in finding the three base numbers, it tells us that, once found, the number a+b+c=n has a perfect cube. Moreover, each number m=hn, with $$h \in \mathbb N$$ is a perfect cube either.
That's it. I don't know where this has been proven, because this is only the result of my spare time calculations. I hope some of you might find it interesting enough to share opinions and ideas.
Now I'm trying the same for the more general rule:
$$\sum_{i=1}^nx_i^n = y^n$$
but it already proved to be nasty!