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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 $x^3=a^3+b^3+c^3$.

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 $x$ there exist integers $a$, $b$, $c$ such that $x^3=a^3+b^3+c^3$.

There are bivariate coprime polynomial parametrizations: https://sites.google.com/site/tpiezas/010

$$(a^4-2ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 b-b^4)^3 = (a^4+a b^3)^3$$

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 $x^3=a^3+b^3+c^3$.

There are bivariate coprime polynomial parametrizations:  https://sites.google.com/site/tpiezas/010

(a^4-2*a*b^3)^3+(a^3*b+b^4)^3+(2*a^3*b-b^4)^3==(a^4+a*b^3)^3

There are bivariate coprime polynomial parametrizations:  https://sites.google.com/site/tpiezas/010

$$(a^4-2ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 b-b^4)^3 = (a^4+a b^3)^3$$