Showing that an element is in the ring of integers Given a number field $K = \mathbb{Q}(\alpha)$ for $\alpha$ a root of $x^3 + x^2 -2x + 8$, show that $\beta = \frac{4}{\alpha} \in \mathscr{O}_K$. 
I'm not looking for a solution to the problem, but rather some help getting started. I know that two methods could be to either write $\beta$ as a linear combination of $\{1, \alpha, \alpha^2\}$ or to find it's minimal polynomial and show that it is in $\mathbb{Z}[x]$, but I am having difficulty doing either of these things in this case. Am I missing something obvious? Is there a better method?
Thank you.
 A: Hint: Let $ f(x) = x^3 + x^2 - 2x - 8 $. We have that 
$$ g(x) = x^3 f(4/x) =  -8x^3 - 8x^2 + 16x + 64 $$
and clearly $ g(4/\alpha) = 0 $.
A further note: to show that a number is an algebraic integer, you do not necessarily have to find its minimal polynomial and show that it has integer coefficients - finding any monic polynomial in $ \mathbf Z[x] $ which has said number as a root is enough.
A: Finding some polynomial for $\beta$ is not that difficult:
$$\beta^3=\frac{64}{\alpha^3},\quad\beta^2=\frac{16\alpha}{\alpha^3},\quad \beta =\frac{4\alpha^2}{\alpha^3},\quad\beta^0=\frac{\alpha^3}{\alpha^3} $$
hence
$$ \beta^3-\beta^2+2\beta+8=\frac{64-16\alpha+8\alpha^2+8\alpha^3}{\alpha^3}=0$$
A: The minimal polynomial of $\beta$ is $x^3-x^2+2x+8$. It is irreducible, monic, and has integer coefficients. Note that we only changed two signs in the minimal polynomial of $\alpha$. One way to find this polynomial is to write $\beta^3+c_1\beta^2+c_2\beta+c_3=0$, and use $\alpha^3+\alpha^2-2\alpha+8=0$ to determine $c_1,c_2,c_3$.
