How does one compute integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt[3]{5})$ 
Integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt[3]{5})$

I was told that it is $\mathbb{Z}[\sqrt[3]{5}]$. I do not find this clear. Is there an easy way to see this? 
 A: Let $D$ be a cube-free integer, not $\pm 1$. Consider an element 
$$x=a + b\sqrt[3]{D}+c \sqrt[3]{D^2}$$ of $\mathbb{Q}(\sqrt[3]{D})$. The Galois transforms of $x$ inside $\mathbb{Q}(\sqrt[3]{D}, \omega)$ are
$$x_0= a + b\sqrt[3]{D}+c \sqrt[3]{D^2}=x\\
x_1=a + b\sqrt[3]{D}\omega+c \sqrt[3]{D^2}\omega^2\\
x_2=a + b\sqrt[3]{D}\omega^2+c \sqrt[3]{D^2}\omega$$
Assume $x$ is an (algebraic) integer. Then $x_1$, $x_2$ are also integer. Summing the above equality we get $x_0+ x_1 + x_2=3a$, an integer. 
Also, $x_0+ \omega^2 x_1 + \omega x_2 = 3 b \sqrt[3]{D}$ and  $x_0+ \omega x_1 + \omega^2 x_2 = 3 c \sqrt[3]{D^2}$ are algebraic integers. 
Raising to cube and using that $D$ is cube-free, we conclude $3b$, $3c$ are also integers. 
The characteristic polynomial of $x$, $(X-x_0)(X-x_1)(X-x_2)$ equals
$$X^3 -3 a X^2 + (3a^2 - 3D b c )X -(a^3+Db^3 + D^2c^3-3D a b c)$$
Now $x$ is an integer if and only if the coefficients of the above polynomial are in $\mathbb{Z}$. 
Assume $a$ is an integer. 
If only one of the $b$, $c$ is not an integer then $a^3+Db^3 + D^2c^3-3D a b c$ will have at least $3$ in the denominator, contradiction. 
If both $b$, $c$ are non-integers then $D$ must be divisible by $3$. 
But then again we get denominators in the same expression, contradiction. Therefore, $b$, $c$ are also integers. 
Assume  $a\not \in \mathbb{Z}$. From $3a^2 - 3D b c$ we conclude that the denominator of $Dbc$ is $9$. 
We get a contradiction right away if $D$ is divisible by $3$, since that would imply the denominator of $bc$ is $27$, not possible.
Otherwise, $a$, $b$, $c$ have denominator $3$. Replacing $x$ with $ x -(p + q \sqrt[3]{D}+ r\sqrt[3]{D^2}$ we only have to look
for possible solutions with $a,b,c= \pm \frac{1}{3}$. Therefore, we have to look for $A$, $B$,$C$ in $\{-1,1\}$ so that 
$$3\mid (1 - D\cdot B C ) \\
27 \mid (A^3 + D B^3 + D^2 C^3 - 3 D A B C)$$
In the case $D = 5$ we do not get any solutions. 
