From this post Where Fermat's Last Theorem fails, we find the nice,
$$(18+17\sqrt2)^3+(18-17\sqrt2)^3=42^3$$
Using this initial solution, an infinite more can be generated using P. Tait's identity,
$$\big(x(y^3 + z^3)\big)^3 + \big(y(-x^3 - z^3)\big)^3 = \big(z(x^3 - y^3)\big)^3$$
which is true if $x^3+y^3=z^3.\,$ For example, the first leads to a second,
$$(707472 + 276119 \sqrt{2})^3 + (707472 - 276119 \sqrt{2})^3 = 1106700^3$$
and so on infinitely. But it is also the case that,
$$(2+\sqrt{-2})^3+(2-\sqrt{-2})^3=(-2)^3$$ $$(121+23\sqrt{-11})^3+(121-23\sqrt{-11})^3=(-88)^3$$
all of which are unique factorization domains UFD ${\bf Q}(\sqrt n)$. So at first I thought it was peculiar to UFDs, but expanding,
$$(a+b\sqrt n)^3+(a-b\sqrt n)^3=c^3$$
we get the simpler,
$$2 a^3 + 6 a b^2 n = c^3\tag{eq.1}$$
After I tested various $n$, it didn't seem to be limited to UFDs. For example,
$$(256+11\sqrt{-41})^3+(256-11\sqrt{-41})^3=296^3$$
Question: Can $2 a^3 + 6 a b^2 n = c^3$ (labeled as eq.1) be turned into an elliptic curve similar to this post so we can easily find $n$ for which eq.1 is solvable? If not, what are the $-100 < n < 100$ such that $(1)$ has a solution? (Is it in the OEIS?)
P.S. We avoid square $n$, since by FLT $x^3+y^3=z^3$, it just yields trivial solutions $xyz = 0$. Or if $n=-3m^2$, then $(a,b,c) = (3m,\,1,\,0)$ so we avoid $abc =0$ as well.
Elliptic curve version.
$2a^3+6ab^2n = c^3$ can be reduced to $v^2 = -12nx^4+6nx$ with $x=a/c,y=b/c, v=6nxy$.
Furthermore, we can tranform $v^2 = -12nx^4+6nx$ to $Y^2 = X^3-432n^3$ with $X=6n/x,Y=6nv/x^2$.
The solutions are derived from one of the generators, so it may not be the smallest.