I need to find the minimal polynomial of $\beta = i + \sqrt[3]{2}$ over $\mathbb{Q}$ using Cayley-Hamilton theorem.
So far I've found that $\{1, \sqrt[3]{2}, \sqrt[3]{4}, i, \sqrt[3]{2}i, \sqrt[3]{4}i\}$ is a basis of $\mathbb{Q}[\sqrt[3]{2},i]$ as a vector space over $\mathbb{Q}$. Also, the matrix representing the multiplication by $\beta$ using this basis is $$ A = \begin{pmatrix} 0& 0& 2&-1& 0& 0\\ 1& 0& 0& 0&-1& 0\\ 0& 1& 0& 0& 0&-1\\ 1& 0& 0& 0& 0& 2\\ 0& 1& 0& 1& 0& 0\\ 0& 0& 1& 0& 1& 0\\ \end{pmatrix} $$
where the element $$\lambda_{1} + \lambda_{2}\sqrt[3]{2} + \lambda_{3}\sqrt[3]{4} + \lambda_{4}i + \lambda_{5}\sqrt[3]{2}i + \lambda_{6}\sqrt[3]{4}i$$ in $\mathbb{Q}[\sqrt[3]{2},i]$ is represented by the vector $$ \begin{pmatrix} \lambda_{1}\\ \lambda_{2}\\ \lambda_{3}\\ \lambda_{4}\\ \lambda_{5}\\ \lambda_{6}\\ \end{pmatrix}. $$
I could't figure out how to use the characteristic polynomial of $A$ ($p(x) = x^{6} + 3x^{4} - 4x^{3} + 3x^{2}$) to find a polynomial $f(x)$ in $\mathbb{Q}[x]$ such that $f(\beta) = 0$.
The charactestic polynomial should be $p(x)=x^6 + 3 x^4 - 4 x^3 + 3 x^2 + 12 x + 5.$ You need to check if $p$ is irreducible. There are many standard algorithm to check if a polynomial is irreducible and you can use them. Otherwise you can use sagemath or any other tools to check as done here SageMath.