Using Cayley-Hamilton theorem to find minimal polynomial I need to find the minimal polynomial of $\beta = i + \sqrt[3]{2}$ over $\mathbb{Q}[x]$ using Cayley-Hamilton theorem. So far I've found $\{1, \sqrt[3]{2}, \sqrt[3]{4}, i, \sqrt[3]{2}i, \sqrt[3]{4}i\}$ is base of $\mathbb{Q}[\sqrt[3]{2},i]$ as a vector space over $\mathbb{Q}$. Also, the matrix representing the multiplication by $\beta$ using the base 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 couldn´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.
