Writing elements in number field w.r.t. a given power basis - is there an easier way? I've done an exercise, and I got the correct answer, but it was a bit labor-intensive, and I wonder if there is an easier way. Here is the problem:

Let $\theta=\sqrt2-\sqrt[3]2$. Write $\sqrt2$ and $\sqrt[3]2$ as rational combinations of $1, \theta, \theta^2, \theta^3, \theta^4$ and $\theta^5$.

So, here's what I did. I guessed a basis: $\{1, \alpha,\alpha^2,\alpha^3, \alpha^4,\alpha^5\}$, for $\alpha=2^{1/6}$. (This seemed right, mostly because every number in sight is pretty clearly in $\Bbb Q(\alpha)$.)
I was able to write each power of $\theta$ in terms of the $\alpha$ basis fairly easily, and I made their coordinates into the columns of a matrix:
$$\begin{bmatrix} 1 & 0 & 2 & -2 & 4 & -40 \\ 0 & 0 & 0 & 6 & 0 & 40 \\ 0 & -1 & 0 & -6 & 2 & -20 \\ 0 & 1 & 0 & 2 & -8 & 4 \\ 0 & 0 & 1 & 0 & 12 & -2 \\ 0 & 0 & -2 & 0 & -8 & 10 \end{bmatrix}$$
I then reasoned that I wanted to write the columns $\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\sim\alpha^2=\sqrt[3]2$, and $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\sim\alpha^3=\sqrt2$ as linear combinations of the columns of that matrix, so I made an augmented matrix, reduced it, and got answers. They're not the prettiest rational numbers, but they're correct.
I wonder if I'm missing an easier method. This seemed indirect, in the way I had to create that $\alpha$-basis, which I just guessed. Is there a better approach to a problem like this? How do I even know at the outset that the desired numbers are in the field $\Bbb Q(\theta)$?
 A: It can be done differently, but I am not sure if it is at all less laborious...
Step 1: Calculate the minimal polynomial of $\theta$ over $\mathbb Q $.
You would start with $\sqrt 2-\theta=\sqrt [3]2$, cube both sides: $2\sqrt 2-6\theta+3\sqrt 2\theta^2-\theta^3=2$, finally write:
$$\sqrt 2=\frac {2+6\theta+\theta^3}{2+3\theta^2} $$
Now square that and you will get some polynomial $\mu $ of 6th degree such that $\mu (\theta) =0$. In what follows, we will not use the fact that it is minimal, so I will leave it without the proof.
Step 2: Calculate $\sqrt 2$: It is already calculated above, as a rational function of $\theta $, but now we have a minimum polynomial, we can invert the denominator ($2+3\theta^2$): use the Euclidean algorithm to find rational polynomials $u,v $ such that $u(x) (2+3x^2)+v(x)\mu (x)=1$, giving you $u (\theta)=\frac {1}{2+3\theta^2} $. Multiply by the numerator ($2+6\theta+\theta^3$) and reduce modulo $\mu $ and you will get $\sqrt 2$ expressed as a rational polynomial over $\theta $.
Step 3: Calculate $\sqrt [3]2=\sqrt 2-\theta$. (Easy.)
