Proving this element is a unit I'm working through a past paper question:

Now assume that $K=\mathbb Q(\theta)$, where $\theta=\sqrt[3]6$. For $a, b\in \mathbb Q$, write down the norm of $a+b\theta \in K$. Factor $2, 3, 5, 7,\theta,\theta– 1,\theta – 2,\theta– 3$ and $\theta + 4$ into prime ideals.
  Show that $$\mu:=\frac{(\theta - 2)^2 (\theta - 3)(\theta - 1)}{\theta(\theta + 4)}$$ is a unit of $K$. By considering the size of $\mu$ as an element of $\mathbb R$, prove that $\mu\ne ±1$.

I am stuck on 2 sections: 
(1) Proving that $\mu$ is a unit. 
To do this I have shown that the norm of $\mu$ is $-1$, so all that remains to do is show that it is an algebraic integer. I am not sure how to do this. It looks like I should be using the previous part of the question; one idea I had was to make all the prime ideals I have found as factors principal, and then correspondingly factor the components of $\mu$ into irreducible elements, but I got stuck doing this and the question doesn't tell us it's a PID. 
(2) Proving that $\mu\ne ±1$. 
Not sure how to do this - I don't know what it means by 'considering the size of $\mu$'.
Any help would be much appreciated! Thank you. 
 A: Unit
Your prime factorisations should show that the ideal generated by $\theta(\theta+4)$ is equal to the ideal generated by $(\theta-2)^2(\theta-3)(\theta-1)$, since these ideals have the same prime factors. Call this ideal $\mathfrak{a}$.
Now, it is not hard to show that $K$ is the field of fractions of $\mathcal{O}_K$ (and we take it as a standard fact), which means that $\mu = \alpha/\beta$ for $\alpha, \beta \in \mathcal{O}_K$, and by definition of $\mu$ we have
$$
\frac{\alpha}{\beta} = \frac{(\theta-2)^2(\theta-3)(\theta-1)}{\theta(\theta+4)}
$$
Clearing denominators,
$$
\alpha \theta(\theta+4) = \beta (\theta-2)^2(\theta-3)(\theta-1)
$$
which gives us the equation of ideals
$$
(\alpha)\mathfrak{a} = (\beta)\mathfrak{a}
$$
so by cancellation $(\alpha) = (\beta)$, which means there is some unit $u$ with $\alpha = u\beta$. Clearly then $\mu = u$, so $\mu$ is a unit. 
Not $\pm 1$
Note that $1 < \theta < 2$. From this, it is clear that $\theta - 3 < 0$, and $(\theta-2)^2, (\theta - 1), \theta, (\theta + 4)>0$, hence $\mu < 0$. Thus, it suffices to show that $\mu \neq -1$. Suppose that $\mu = -1$. Then we have
$$
(\theta-2)^2(3-\theta)(\theta-1) = \theta(\theta+4)
$$
Again using $1 < \theta < 2$ we have that $(\theta - 2)^2 < 1^1 = 1, (3 - \theta) < 2, (\theta - 1) < 1$, so since these quantities are all positive, we have $(\theta-2)^2(3-\theta)(\theta-1) < 2$. On the other hand, we have $\theta>1$ and $\theta + 4 > 5$, so $\theta(\theta+4) > 5$. Clearly then these cannot be equal, and hence $\mu \neq -1$.
