Ideal class group of $\mathbb{Q}(\sqrt[3]{5})$

Question

I want to calculate the ideal class group of $K=\mathbb{Q}(\sqrt[3]{5})$
The ring of integers $O_K=\mathbb{Z}[\sqrt[3]{5}]$, discriminant $d_K=-27\cdot5^2=-675$, Minkowski constant $M_K=\frac{3!}{3^3}(\frac{4}{\pi})^1\sqrt{675}<8$
Then $Cl(K)$ is generated by ideals $I$ s.t. $N(I)\leq7$

Consider factorization of $2,3,5,7$ in $O_K$, we see that:
$(2)=(2,\sqrt[3]{5}+1)(2,\sqrt[3]{25}-\sqrt[3]{5}+1)$, $(3)=(3,\sqrt[3]{5}+1)^3$, $(5)=(\sqrt[3]{5})^3$, and $(7)$ is prime in $O_K$

Therefore the class group is generated by $I_2=(2,\sqrt[3]{5}+1),I_3=(3,\sqrt[3]{5}+1)$
$I_3^3=(3)$ is principal, $I_2I_3=(\sqrt[3]{5}+1)$ is principal
So the class group is generated by $I_3$, whose order is 1 or 3

I guess that $I_3$ isn't principal, but I didn't find a good way to prove it.

If $I_3=(\alpha)=(a+b\sqrt[3]{5}+c\sqrt[3]{25})$, then $N(\alpha)=3$, i.e. $a^3+5b^3+25c^3-15abc=3$
I don't know how to treat this equation.

Answer 1

Ohh, I found a solution! Use GeoGebra, I draw the diagram for $y=\sqrt[3]{5}x$ and $y=x/\sqrt[3]{5}$
I found that the points $(7,4),(7,12)$ are very near to the two lines.
After calculation, I found that $12+7\sqrt[3]{5}+4\sqrt[3]{25}$ has norm 3,
$3=(12+7\sqrt[3]{5}+4\sqrt[3]{25})(4-4\sqrt[3]{5}+\sqrt[3]{25})$
$1+\sqrt[3]{5}=(12+7\sqrt[3]{5}+4\sqrt[3]{25})(3-\sqrt[3]{25})$

Therefore $I_3$ is principal, generated by $12+7\sqrt[3]{5}+4\sqrt[3]{25}$