Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$ It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
Accordingly, I have been looking into instances where $c^3|(a^3+b^3)$ for $a,b\in\mathbb{N}$. In other words, solutions to the Diophantine Equation:
$a^3+b^3=dc^3$ where $a,b,$ and $c$ are pairwise co-prime and $a,b,c>0$
Obviously there are some trivial solutions. If $c=1$, for example, $a$ and $b$ can be any integers, and $d$ can be chosen to simply be $a^3+b^3$. 
By requiring that $a,b,$ and $c$ are pairwise co-prime and that $c\not=1$, we eliminate the trivial solutions, and what remains is of much more interest. 
For $a,b,c,d\le20$, there are 5 solutions:
$4^3+5^3=7*3^3$
$2^3+7^3=13*3^3$
$1^3+8^3=19*3^3$
$3^3+5^3=19*2^3$
$1^3+19^3=20*7^3$
Under $100$ there are $16$ solutions, as found by Mathematica.
My  question about this equation: Has it been studied previously? Are there infinitely many primitive solutions (which it seems like there are)? If so, can they be parametrized?

There are infinitely many parametric solutions for arbitrary c.
We use simple identity below.
$$(c^3n+c^3-b)^3+b^3=c^3(n+1)(c^6n^2+2c^6n-3c^3nb+3b^2+c^6-3c^3b)$$
Let $a=c^3n+c^3-b, d=(n+1)(c^6n^2+2c^6n-3c^3nb+3b^2+c^6-3c^3b)$.
Hence we get the parametric solutions of $a^3+b^3=dc^3$.
We show the examples only for $c=2,3,4,5.$ 
$$(8n+7)^3+1^3 = 2^3(n+1)(64n^2+104n+43)$$
$$(27n+26)^3+1^3 = 3^3(3n+3)(243n^2+459n+217)$$
$$(64n+61)^3+3^3 = 4^3(n+1)(4096n^2+7616n+3547)$$
$$(125n+124)^3+1^3 = 5^3(n+1)(15625n^2+30875n+15253)$$
