Zero sum of four cubes in K[X]

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Good evening,

I'm trying to solve the equation

$A^3+B^3+C^3+D^3=0$

in the ring of complex polynomials; indeed, I don't know if I can find all solutions, prove there aren't non constant solutions or exhibit one.

One should certainly add a condition such as $\gcd(A,B,C,D)=1$ in order to avoid other kind of trivial solutions like $(A,B,-A,-B)$.

I tried to use the Mason-Stothers theorem, since it may be used to show Fermat's Last Theorem for polynomials or Catalan's conjecture for rational functions, but I could not get a strong enough condition on degrees to get a solution or a contradiction.

May anybody help me?

Thanks!

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Examples can be found based on known parametrizations of the solutions to the diophantine equation $A^3+B^3=C^3+D^3\,$ (see 3.2.2 here). For example, Ramanjuan's identity:

$$(a^2+7ab-9b^2)^3+(2a^2-4ab+12b^2)^3=(2a^2+10b^2)^3+(a^2-9ab-b^2)^3$$

For $b=1$ the above gives:

$$(x^2+7x-9)^3+(2x^2-4x+12)^3+(-2x^2-10)^3+(-x^2+9x+1)^3 = 0$$

It can be easily verified that the $4$ polynomials are mutually coprime.