The minimal poly of $\sqrt[3]{2}$ over $\Bbb{Q}$ is equal to $\det(T_a - xI)$ where $T_a$ is a matrix over $\Bbb{Q}$ that represents mult. by $a$. Let $K/F$ be a field extension of degree $n \in \Bbb{N}$ and for each $a \in K$ define $L_a(x) = a x$.  Then $L_a(x)$ is an $F$-linear transformation of $K$ as a vector space of dimension $n$.  So send $K$ into $F^{n \times n}$ the matrix ring by sending $a$ to $T_a = [ L_a(\theta_1) \ \cdots \ L_a(\theta_n) ]$ where abstractly we have $L = \{ a_1 \theta_1 + \dots + a_n \theta_n : a_i \in F\}$ for some $\theta_i$ basis in $K$.
Then for $a \in K$, $f(x) = \det (T_a - xI) \in F[X]$ the characteristic polynomial, we have that $f(a) = 0$ i.e. that $a$ is a root of the characteristic polynomial which is monic of degree $n$ so is the characteristic polynomial is in fact $m_{a, F}(x)$ the minimal polynomial for $a$ over $F$.
I'm trying to prove this in the general case, i.e. that $f(a) = 0$ or equivalently that $T_a(y) = ay$ for all $y \in F^n$.
What I have so far is:
$$
T_a(y) = \sum_{i=1}^n y_i L_a(\theta_i) \\
\implies T_a(y) = \sum_{i=1}^n y_i (a \theta_i) = a \cdot \dots
$$
So I've got that so far.  Then the problem says, test this idea out to find the monic of degree $3$ satisfied by $a = \sqrt[3]{2}$.
So I want to compute the determinant of:
$$
\begin{pmatrix}
x - \sqrt[3]{2} & 0 & 0 \\
0 & x - \sqrt[3]{4} & 0 \\
0 & 0 & x - 2 
\end{pmatrix}
$$
where I've reversed the sign for simplicity.  I computed the above by multiplying $\theta_1 = 1, \theta_2 = \sqrt[3]{2}, $ and $\theta_3 = \sqrt[3]{4}$ by $a$ and subtracting that from $x$.
I'm getting:
$$
x^3 - 2 x^2 + (2 + 2\sqrt[3]{2} + 2\sqrt[3]{4}) x - 4
$$
which is not a polynomial over $F$.  The bad term I got by doing $(-\sqrt[3]{2})(-2) + (-\sqrt[3]{4})(-2) + (-\sqrt[3]{2})(-\sqrt[3]{4})$ in a logical, symmetric way.
Where have I gone wrong in my computation?

I think you've computed $T_a$ incorrectly. I assume you're using the ordered basis $(1,\sqrt[3]{2},\sqrt[3]{4})$ for $K$ as $\mathbb Q$-vector space (edit: I see that you are). So applying $L_a$ to the first basis vector gives $L_a(1)=\sqrt[3]{2}$. In terms of the coordinate vectors relative to this ordered basis, this is $$L_a\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix}$$ so this should be the first column of $T_a$. I'll leave it to you to check the other columns.
Lastly, a small note: it should be $\det(xI-T_a)$ for the minimal polynomial to be monic.
