Two Sums of Three Cubes

Note that:

$(a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3a^2c + 3ac^2 + 3b^2c + 3bc^2 + 6abc$

$(a + b + c)^3 + a^3 + b^3 + c^3 = a^3 + 3a^2b + 3ab^2 + b^3 + a^3 + 3a^2c + 3ac^2 + c^3 + b^3 + 3b^2c + 3bc^2 + c^3 + 6abc$

$(a + b + c)^3 + a^3 + b^3 + c^3 = (a + b)^3 + (a + c)^3 + (b + c)^3 + 6abc$

If $c^3 = 6abc$ , then $c^2 = 6ab$ , and:

$(a + b + c)^3 + a^3 + b^3 = (a + b)^3 + (a + c)^3 + (b + c)^3$

Setting $a = 2$ and $b = 3k^2$ gives $c = 6k$ and the generator:

$(3k^2 + 6k + 2)^3 + 2^3 + (3k^2)^3 = (3k^2 + 2)^3 + (6k + 2)^3 + (3k^2 + 6k)^3$

which produces infinitely many solutions that follow the given constraints when $k$ is a positive odd integer, such as:

$2^3 + 3^3 + 11^3 = 5^3 + 8^3 + 9^3$

$2^3 + 27^3 + 47^3 = 20^3 + 29^3 + 45^3$

$2^3 + 75^3 + 107^3 = 32^3 + 77^3 + 105^3$

and so on.

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Some similar generators can be found. Setting $a = 6$ and $b = k^2$ gives $c = 6k$ and the generator:

$(k^2 + 6k + 6)^3 + 6^3 + (k^2)^3 = (k^2 + 6k)^3 + (6k + 6)^3 + (k^2 + 6)^3$

and setting $a = 3$ and $b = 2k^2$ gives $c = 6k$ and the generator:

$(2k^2 + 6k + 3)^3 + 3^3 + (2k^2)^3 = (2k^2 + 6k)^3 + (6k + 3)^3 + (2k^2 + 3)^3$

Thanks to Hardy's cab number and Ramanujan we know that $9^3+10^3+r^3=1^3+12^3+u^3$ will hold for any positive integer $r=u$ . Also note that $\gcd(p,q,s,t)=1 \implies \gcd(p,q,r,s,t,u)=1$ So we already found an infinite subset of the solutions.

Edit: The values had to be different, I did not read it well...