Step in a solution of $y^2 = x^3 - 2$ I am reading Algebraic Number Theory notes here by Keith Conrad. In page 9, there is a solution of $y^2=x^3-2$ using unique factorization in $\mathbb{Z}[\sqrt{-2}]$. We start by writing $x^3=y^2+2=(y+\sqrt{-2})(y-\sqrt{-2})$. By some work, it is shown that $y$ is odd, and that only common divisors of $y+\sqrt{-2}$ and $y-\sqrt{-2}$ are $\pm 1$. My question is about the following step:

By unique factorization in $\mathbb{Z}(\sqrt{-2})$,
  $y+\sqrt{-2}=u\alpha^3$ where $u$ is a unit.

Why does this step follow? Well, intuitively because $(y+\sqrt{-2})(y-\sqrt{-2})=x^3$ is perfect cube, and the factors $y+\sqrt{-2}$ and $y+\sqrt{-2}$ have no common divisors (other than $\pm 1$), so each factor must be a cube on its own (up to some unit). But how do we rigorously prove what I just said? 
Thanks for your time.
 A: Thanks to the encouragement from Sanchez, I have found the answer (making it community wiki):
Suppose $p$ is any prime dividing $y+\sqrt{-2}$. Then, $p$ divides $x^3$, and since $p$ is prime, $p$ divides $x$. Thus, 

$p^3$ divides $x^3=(y+\sqrt{-2})(y-\sqrt{-2})$     (*)            

Now, $p$ divides $y+\sqrt{-2}$. Also, $y+\sqrt{-2}$ and $y-\sqrt{-2}$ are relatively prime. So, $p$ cannot divide $y-\sqrt{-2}$. Since $p$ is prime, $p$ must be relatively prime to $y-\sqrt{-2}$, and consequently, $p^3$ is relatively prime to $y-\sqrt{-2}$. Using (*), we get that $p^3$ divides $y+\sqrt{-2}$, and as a result $y+\sqrt{-2}$ is a cube (up to a unit), as desired.
Here I have used two facts concerning UFDs:
1) If $a$ is relatively prime to $b$, then $a^3$ is relatively prime to $b$. 
2) If $a$ divides $bc$, and furthermore, $a$ is relatively prime to $b$, then $a$ must divide $c$.
A: Here's two slightly different ways to think about it that you might prefer, I'll write them out in full generality just because.
Suppose, in a unique factorisation domain $D$ we have the following equation:
$$x^n=ab$$
If $(a,b)=1$ (up to units) then both $a$ and $b$ are perfect $n^{th}$ powers multiplied by units.
Method 1: We show by induction that if $p|a$, $p^k|a$ for any integer $k\leq n$. (From this the result is clear by writing out $a$ in terms of its irreducible factors and using that in a UFD all irreducibles are prime, or by iterating the process on $\frac{a}{p^n}$ and using the fact that $a$ has only finitely many prime factors.)
Suppose $p|a$. This gives the case $k=1$ so all that remains is the inductive step, so suppose now that $k\leq n$ and $p^{k-1} | a$. We already know that $p|x^n$, so $p$ divides $x$. Dividing both sides by $p^{k-1}$ gives:
$$\bigg(\frac{x}{p}\bigg)^{k-1}x^{n-k+1}=\frac{a}{p^{k-1}}b$$
$p |x^{n-k+1}$ so either $p|\frac{a}{p^{k-1}}$ or $p|b$. The latter is impossible, so we must have that $p|\frac{a}{p^{k-1}}$ i.e. $p^k |a$.
Method 2: We use unique factorisation to prove the statement directly. (Sorry for the mess of indices!)
As we are in a unique factorisation domain, we have:
$$x=u_x x_1^{k_1}\dots x_{n_x}^{k_{n_x}}, a=u_a a_1^{l_1}\dots a_{n_a}^{l_{n_a}},b=u_b b_1^{m_1}\dots x_{n_b}^{m_{n_b}}$$
Where $u_x,u_a,u_b$ are all units, $x_i\not\sim x_j,a_i\not\sim a_j,b_i\not\sim b_j$ if $i\not=j$ and $a_i\not\sim b_j$ for any $i,j$ (as $a,b$ are coprime). Plugging this in gives:
$$u_x^n x_1^{nk_1}\dots x_{n_x}^{nk_{n_x}}=u_au_b a_1^{l_1}\dots a_{n_a}^{l_{n_a}} b_1^{m_1}\dots x_{n_b}^{m_{n_b}}$$
By uniqueness, it follows that $n_x=n_a+n_b$ and after re-arrangement and multiplication by units if necessary, $x_1=a_1,\dots ,x_{n_a}=a_{n_a},x_{n_a+1}=b_1,\dots  x_{n_a+n_b}=b_{n_b}$ and also $nk_1=l_1,\dots,nk_{n_a}=l_{n_a},nk_{n_a+1}=m_1,\dots,nk_{n_a+n_b}=m_{n_b}$. The result should now be clear.
