Let $K=\mathbb Q\left(^3\sqrt{5}\right)$ and $\alpha=a+b\left(^3\sqrt{5}\right)+c\left(^3\sqrt{5}\right)^2$. How do I find the minimal polynomial $f_\alpha$ of $\alpha$ over $\mathbb Q$?
I am already aware of the method that involves using a matrix of $m_\alpha$, I want to be able to find the polynomial without use of matrices.
I have tried cubing $\alpha$ in an attempt to multiply out the cube root terms, however this gets messy quickly and creates as many terms in powers of $^3\sqrt{5}$ as it eliminates. Does the fact that the minimal polynomial of $^3\sqrt{5}$ over $\mathbb Q$ is $x^3-5$ help here since $\alpha \in K=\mathbb Q\left(^3\sqrt{5}\right)$?
The reason I want to find the minimal polynomial is to compute the Norm $N_{K/\mathbb Q}(\alpha)$ and the Trace $Tr_{K/\mathbb Q}(\alpha)$ from it, using its conjugates. Is there an easier way to do this (again, without using matrices)?
When you know the conjugates of the element in question, you can write out the Norm, Trace, and characteristic polynomial without resorting to matrices. For instance, in the case you mentioned, $\xi=a+b\lambda+c\lambda^2$, where $\lambda^3=5$, we have $\xi'=a+b\omega\lambda+c\omega^2\lambda^2$ and $\xi''=a+b\omega^2\lambda+c\omega\lambda^2$, for the conjugates of $\xi$. (Here, of course, $\omega$ is a primitive cube root of unity, so a root of the polynomial $X^2+X+1$.) Then the Trace is $\xi+\xi'+\xi''$, the Norm is $\xi\xi'\xi''$, and the characteristic equation is $(X-\xi)(X-\xi')(X-\xi'')$. All of these are computable by hand, and the exercise of working slowly and checking each step is very valuable.