Sum of three perfect cubes is equal to a perfect fourth How many answers does the following equation have?
$a^3+b^3+c^3=d^4$ where $a, b, c, d\in \mathbb{Z}^+$.
This was asked in a test for gifted math students in 7th grade in Finland. I have been thinking about this and couldn't solve this for my number theory isn't that good.
Could someone help me? Thank you in advance.

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
