There are bivariate coprime polynomial parametrizations: https://sites.google.com/site/tpiezas/010
$$(a^4-2ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 b-b^4)^3 = (a^4+a b^3)^3$$
Added If you drop the positivity constraint then there is another identity for $x=v^4$.
$$v^{12}=(v^4)^3=(9u^4)^3+(3uv^3-9u^4)^3+(v(v^3-9u^3))^3$$
I believe, but can't find it at the moment, that for all positive $x$ exist integers $a,b,c$ such that $x^3=a^3+b^3+c^3$.