If $x = y^3$ in $\mathbb{Z} + \frac{-1 + \sqrt{-3}}{2}\mathbb{Z}$, then there is some $w \in \mathbb{Z} + \sqrt{-3}\mathbb{Z}$ such that $x = w^3$. So the title contains the question I have to solve:
"If $x \in \mathbb{Z} + \frac{\bf{\color{red}{-}}1 + \sqrt{-3}}{2}\mathbb{Z}$ such that it can be written as the third power of something in $\mathbb{Z} + \frac{\bf{\color{red}{-}}1 + \sqrt{-3}}{2}\mathbb{Z}$, then it can be written as the third power of something in $\mathbb{Z} + \sqrt{-3}\mathbb{Z}$."
I was able to show that if $R$ is a UFD and $a,b,c \in R$ such that $ab = c^p$ with $p > 1$ some integer and if $g = \text{gcd}(a,b)$, then there exists $d, x \in R$ such that $d \mid g^{p-1}$ and $a = dx^p$. The follow up question was the above, but I have no idea how to start this question: I have tried brute force, i.e. say $x = a + b\phi$, $y = c + d\phi$ with $\phi = \frac{\bf{\color{red}{-}}1 + \sqrt{-3}}{2}$ and then I have tried to show that $d$ would need to be an even number. However, this approach was not really helpful. Moreover, I suspect that I would need the above, but I do not see how. (I know that $\mathbb{Z} + \phi\mathbb{Z}$ is a UFD, whereas $\mathbb{Z} + \sqrt{-3}\mathbb{Z}$ is not).
any hints?
Remark I have initially made a mistake in the sign of $1$ in the fraction, I edited this in red.
 A: Let $\phi$ be a third root of unity, i.e. $\phi^3=1$ and $\phi^2+\phi+1=0$.
The trick is the following:

Lemma. If we are given some element $y=a+b\phi \in \mathbb Z+\phi\mathbb Z$,
  then one of the three elements $y,y\phi,y\phi^2$ is contained in
  $\mathbb Z+\sqrt{-3}\mathbb Z$.

Proof: We have $\phi^2=-(\phi+1)$, i.e. $y\phi^2=-(y\phi + y)$. The sum of two elements in $(\mathbb Z+\phi\mathbb Z)\setminus (\mathbb Z+\sqrt{-3}\mathbb Z)$ is clearly in $\mathbb Z+\sqrt{-3}\mathbb Z$. So if both summands on the RHS are not contained in $\mathbb Z+\sqrt{-3}\mathbb Z$, then the LHS is. $\small\Box$
Furthermore we have $\phi^3=1$, i.e. $$y^3=(y\phi^2)^3=(y\phi)^3.$$ Thus one of those three guys is the $w$ you search for.

Note that this solution is somehow a priori clear, there is no chance to avoid this Lemma: If $x=y^3=w^3$, we have that $\frac{w}{y}$ is a third root of unity, i.e. $w \in \{y,y\phi,y\phi^2\}$. So the truth of the Lemma I proved was the only chance we had to begin with.
A: Suppose $x=y^3$, where $y \in \mathbb{Z} + \phi\mathbb{Z}$, with $\phi = \displaystyle{\frac{-1+\sqrt{-3}}{2}}$.

Then either 
\begin{align*}
\;y &\in \mathbb{Z} + \sqrt{-3}\mathbb{Z}\\[4pt]
&\qquad\;\;\text{or}\\[4pt]
\;y &= \frac{a + b\sqrt{-3}}{2},\;\text{where $a,b$ are odd integers}\\[4pt]
\end{align*}
In the first case, just let $w=y$, and we're done.

Thus, assume the second case.

Since $a,b$ are odd, either
$$a \equiv b\;(\text{mod}\,4)\;\;\text{or}\;\;a \equiv -b\;(\text{mod}\,4)
\qquad\qquad\;\,$$
If $a \equiv b\;(\text{mod}\,4),\;$let $w = \phi\,y$.
\begin{align*}
\text{Then}\;\;w &= \phi\, y\\[6pt]
&= \left(\frac{-1 + \sqrt{-3}}{2}\right)\left(\frac{a + b\sqrt{-3}}{2}\right)\\[6pt]
&= \left(-\frac{a+3b}{4}\right) + \left(\frac{a-b}{4}\right)\sqrt{-3}\\[6pt]
&\in\, \mathbb{Z} + \sqrt{-3}\mathbb{Z}\\[8pt]
\text{and}\;\;w^3 &= (\phi y)^3 = \phi^3y^3 = (1)(y^3) = y^3 = x,\;\text{as required}\\[4pt]
\end{align*}
If $a \equiv -b\;(\text{mod}\,4),\;$let $w = \phi^2 y$.
\begin{align*}
\text{Then}\;\;w &= \phi^2 y\\[6pt]
&= \left(\frac{-1 - \sqrt{-3}}{2}\right)\left(\frac{a + b\sqrt{-3}}{2}\right)\\[6pt]
&= \left(-\frac{3b-a}{4}\right) + \left(-\frac{a+b}{4}\right)\sqrt{-3}\\[6pt]
&\in\, \mathbb{Z} + \sqrt{-3}\mathbb{Z}\\[8pt]
\text{and}\;\;w^3 &= (\phi^2 y)^3 = (\phi^2)^3y^3 = (1)(y^3) = y^3 = x,\;\text{as required}\\[4pt]
\end{align*}
Thus, in all cases, the specified requirements are met.

This completes the proof.
