$\color{red}{\text{I forgot a vital condition on the integers $a,b$: they must be coprime!}}$
Consider $\omega = \frac{-1-\sqrt{-3}}{2}$, then the ring $\mathbb{Z}+\omega\mathbb{Z}$ is known to be a UFD (unique factorization domain). $\color{red}{\text{Suppose that $a,b$ are coprime integers. }}$ I was able to show that $a + b\sqrt{-3}$ and $a - b\sqrt{-3}$ ($a,b$ integers) only have units as common divisors. Moreover, it is given that $a^2+3b^2 = c^3$ for an integer $c$. From this, I found that $$(a + b\sqrt{-3})(a-b\sqrt{-3}) = c^3$$ so looking at the unique factorization, I find that $$a+b\sqrt{-3} = u(d + e\omega)^3$$ for some integers $d,e$ and $u$ a unit. I know the units in $\mathbb{Z}+\omega\mathbb{Z}$ are given by $\{\pm1, \pm\omega, \pm\omega^2\}$. Because of the structure of the exercise which was given, I know I should be able to show that $a + b\sqrt{-3}$ is a cube in $\mathbb{Z}+\omega\mathbb{Z}$, which I an show if the unit would be a cube itself. However, I am stuck and don't know how to proceed. Any hints?
edit: Perhaps I should give some more details on the exercise: $\color{red}{\text{The purpose is to show that } a + b\sqrt{-3} \text{ can be written as} (d + e\sqrt{-3})^3}$. In class, we saw that $\mathbb{Z}+\omega\mathbb{Z}$ is an Euclidean ring. The exercise consists out of multiple parts:
In part 1, we have shown that in a UFD it holds that if $ab = c^p$ (for $a,b,c$ elements in this UFD) for some integer $p$ larger or equal to $2$ and $g = \text{gcd}(a,b)$, then there exists a unit $d$ and an element $x$ in the UFD such that $a = dx^p$ en $d \mid g^{p-1}$.
In part 2, we have shown that if $x$ is a cube in $\mathbb{Z} + \omega\mathbb{Z}$, then it is in $\mathbb{Z} + \sqrt{-3}\mathbb{Z}$.
Part $3$ is the question where I have trouble with. From part $1$, I know that $$a + b\sqrt{-3} = dx^3$$ with $d$ a unit (since $\text{gcd}(a+b\sqrt{-3}, a-b\sqrt{-3})$ is a unit). Now if I could show that $d$ is a cube itself, then I coul go on.
Let $a+b\sqrt{-3}=up_1^{m_1}p_2^{m_2}\dotsm p_r^{m_r}$ and $a-b\sqrt{-3}=vq_1^{n_1}q_2^{n_2}\dotsm q_s^{n_s}$, where $p_i$ and $q_j$ are primes in your ring (with $p_i$ and $q_j$ not associate), $u$ and $v$ are units.
Then $$ c^3=(a+b\sqrt{-3})(a-b\sqrt{-3})= uvp_1^{m_1}p_2^{m_2}\dotsm p_r^{m_r} q_1^{n_1}q_2^{n_2}\dotsm q_s^{n_s} $$ This is a factorization into primes, so the integers $m_i$ and $n_j$ are all multiples of $3$, by uniqueness of the factorization. Therefore $$ a+b\sqrt{-3}=ux^3 $$ for some $x$ in the ring and $u$ a unit.
This cannot be reduced to $a+b\sqrt{-3}$ being a cube. Consider $$ (2+2\omega)^3=-8 $$ Then $$ \omega(2+2\omega)^3=-8\omega=4+4\sqrt{-3} $$ is of the form $a+b\sqrt{-3}$ with $a$ and $b$ integers, but is not a cube, because $\omega$ is not a cube.