Question related to an unramified cubic extension polynomial

Hello

I have a question regarding the computation of the unramified cubic extensions of the real quadratic field K_D with discriminant D=4*(m)^3-27*(n)^2= 48035713, for m=229, n=3. I wrote the discriminant like this since in cases where the class number is divisible by 3 the discriminants can be written in this form, for finitely many pairs (m,n), and the polynomials f=x^3-m*x+n define unramified cubic extensions (when gcd(2m,3n)=1).

The ideal class group CL(K_D) of K_D has 3-rank 2 and is of the form (C6)x(C6). I computed the four polynomials giving me the 4=(3^2-1)/2 unramified cubic extensions but only one of the polynomials has poldisc equal to the discriminant of K_D, namely the polynomial f. The following is a classical result of Hasse:

"For K a quadratic number field of discriminant D, and r the 3-rank of the ideal class group of K, the number of non-conjugate cubic fields of discriminant D is m = (3r − 1)/2."

My question is two-fold: In this case for this D= 48035713, is it because of computational limitations that PARI gives me only one polynomial with discriminant D (namely the f=x^3-mx+n)?

Finding another polynomial of discriminant D would give a specific elliptic curve, namely the curve E:y^2=x^3-27*16*D, a second point (the first one is the point P=[12m,108n]) which would be independent form P and this would result to the curve having rank>=2 instead of rank=1 which is what I get now.

I hope you can answer this question for me.

Thank you for taking the time to read my e-mail!

Best regards

Eleni Agathocleous