Sum of three perfect cubes is equal to a perfect fourth

Note that if $a^3 + b^3 + c^3 = n$, then $(na)^3 + (nb)^3 + (nc)^3 = n^4$. So you have infinitely many solutions even if you require $a,b,c$ to be distinct.

There are also solutions where $a,b,c$ are pairwise coprime, e.g.

$$ \eqalign{19^3 + 89^3 + 117^3 &= 39^4\cr 107^3 + 163^3 + 171^3 &= 57^4\cr 81^3 + 167^3 + 266^3 &= 70^4\cr 75^3 + 164^3 + 293^3 &= 74^4} $$

Are there infinitely many of those?

[EDIT] See OEIS sequence A327586

It has an infinite number of solutions. For instance, for any $k\in \mathbb{N}$, take $d=3k^3$, and $a=b=c=3k^4$.

Then $$ a^3+b^3+c^3=3\cdot 3^3k^{12} = 3^4k^{12}=d^4.$$