Implications of non unique factorization in diophantine equations I am after a concrete example in which under the below conditions, $x+k\sqrt{-d}$ in $Z[-\sqrt{d}]$ is not a cube or an associate of a cube.
Suppose that $Z[-\sqrt{d}]$ ($d$ squarefree) is a non Unique Factorization Domain (UFD). Consider the equation $x^2+k^2d=y^3$, which factors into $(x+k\sqrt{-d})(x-k\sqrt{-d})$ in $Z[-\sqrt{d}]$.
Suppose further that $x+k\sqrt{-d}$ and $x-k\sqrt{-d}$ have no common factor in $Z[-\sqrt{d}]$. 
Since $Z[-\sqrt{d}]$ is not UFD, in general we cannot assume that $x+k\sqrt{-d}$ is a cube or an associate of a cube. I am looking for a concrete example of this last statement. 
I have left the equation $x^2+k^2d=y^3$ quite open to facilitate the search for an example.
Any help is greatly appreciated...
Thanks,
Guille
 A: I am not completely sure that I understand what you are asking for, but here's an attempt: the phenomenon you describe has implications for the solution set of the Mordell equation $y^2 + k = x^3$ (this is the same Diophantine equation you asked about, with slightly different notation).  This is treated in this exposition of Keith Conrad and this exposition of mine.
Here is Proposition 4 from my link, which seems especially relevant:
Proposition: Let $R$ be a UFD, $n \in \mathbb{Z}^+$ and $x,y,z \in R$ be coprime elements such that $xy = z^n$.
a) Then there exist $\alpha,\beta \in R$ and units $u,v \in R^{\times}$ such that $x = u \alpha^n, y = v \beta^n$.
b) If every unit in $R$ is an $n$th power, then there exist $\alpha,\beta \in R$ such that $x = \alpha^n$, $y = \beta^n$.
In Section 4 from my link, it is shown that when the hypothesis of this proposition  applies to $R = \mathbb{Z}[\sqrt{-k}]$ (or actually a weaker condition, which amounts to the quadratic order $\mathbb{Z}[\sqrt{-k}]$ being maximal and of class number prime to $3$) the Mordell equation has very few solutions $(x,y) \in \mathbb{Z}^2$, and all solutions can be explicitly described.  Contrapositively, examples are given where the Mordell equation has "unexpected" extra solutions, which are in fact concrete examples of when the assumption you are asking about is not satisfied.
Added: Let me reproduce a little bit more from my notes.
Theorem: Suppose $k$ is a squarefree positive integer with $k \equiv 1,2 \pmod 4$.  Suppose the quadratic ring $R = \mathbb{Z}[\sqrt{-k}]$ has property "CM(3)": for all $x,y,z \in R$ such that the ideal generated by $x$ and $y$ is all of $R$ and $xy = z^3$, then $x$ and $y$ are cubes in $R$ up to units: there exist $\alpha,\beta \in R$ and $u,v \in R^{\times}$ such that $x = u \alpha^3$, $y = v \beta^3$.  [In particular this holds when $R$ is a UFD and in fact whenever its class number is prime to $3$.  Note that the conditions on $k$ ensure that $R$ is the full ring of integers in $\mathbb{Q}(\sqrt{-k})$.]  Then the Mordell equation $y^2 + k = x^3$ has either no integer solutions or two integer solutions.  [If you read the handout you will find an explicit condition in terms of $k$ which tells you which case you are in, and also exactly what the two solutions are if you are in that case.]
Example: Take $k = 26$.  Then the equation $y^2 + 26 = x^3$ has at least $4$ solutions: $(x,y) = (3,\pm 1), (35,\pm 207)$, so the ring $\mathbb{Z}[\sqrt{-26}]$ does not satisfy property CM(3).  This is an explicit example of what (I think) the OP is looking for.  Well, I am not being completely explicit about what the $x,y,z$ are in this case, but it is a good exercise to track through the proofs and find them!
A: $-64+49\sqrt{-23}=(37+6\sqrt{-23})(2+\sqrt{-23})$ is (if I've done my calculations right) not a cube, not an associate of a cube, not irreducible, and has no common factor with $-64-49\sqrt{-23}$, but $(-64)^2+(23)(49)^2=39^3$. [The part about "not irreducible" is in answer to OP's request for same in a comment on Pete Cla
