It is well-known that, $$a^3+b^3+c^3 = N\tag1$$ for $N=1,\,2$ is solvable in the integers in infinitely many ways . However, it is an open question (but is conjectured) that if for general $N$ it has a solution, then it must have infinitely many. I'd like to propose a variant,
Conjecture: If $a^3+b^3+c^3=N$, then $x^3+y^3+z^3+t^3 = N$ in infinitely many ways.
Part I: Use the identity,
$$x^3+y^3+z^3+t^3 = (a^3+b^3+c^3)\big(p^2+(a+b)cq^2\big)^3$$
where,
$$\begin{aligned} t &= a p^2 - c^2 p q + b (a + b) c q^2 \\ x &= b p^2 + c^2 p q + a (a + b) c q^2\\ y &= c p^2 + (a^2 - b^2) p q\\ z &= -(a^2 - b^2) p q + (a + b) c^2 q^2\end{aligned}$$
Part II: Solve the Pell equation,
$$p^2+(a+b)cq^2 = 1\tag2$$
This is the tricky part since we require $(a+b)c<0$. Perusing through S. Huisman $15254$ solutions, I find that all his solvable $N$ have solutions where both $a,b$ are negative, or $c$ is negative. Of course, $(2)$ can then be easily solved as a Pell equation.
Q: If $(1)$ has a solution in positive integers, then how do we prove that implies a subsequent solution where one or two of the variables are negative?
P.S. The second solution may be huge, $$4^3+ 4^3+ 7^3 =-17863863023^3+ 13242033973^3+ 15005424841^3 = 471$$ but it guarantees infinitely many solutions to $(2)$.
