If $\alpha = \sqrt 2 + \sqrt 3$, it is easy to find $\sqrt 2$ in terms of $\alpha$: simply invert $\alpha$ to obtain $\sqrt 3 - \sqrt 2$, hence $\sqrt 2 = (\alpha - 1/\alpha)/2$.
What about if $\beta = \sqrt 2 + \sqrt[3]3$? I tried computing powers of $\beta$ (including $1/\beta$) but it didn't seem to go anywhere.
Here is how to express $\sqrt[3]3$ in terms of $\gamma = \sqrt[3]3 + \sqrt[5]5$.
First, note that
$$5 = (\gamma - \sqrt[3]3)^5 = \gamma^5 - 5\sqrt[3]3\gamma^4 + 10\sqrt[3]9\gamma^3 - 30\gamma^2 + 15\sqrt[3]3\gamma - 3\sqrt[3]9.$$
Let $\gamma_1 = \frac{15 - 5\gamma^4}{\gamma^5 - 30\gamma^2 - 5}$, $\gamma_2 = \frac{10\gamma^3 - 3}{\gamma^5 - 30\gamma^2 - 5}$.
Then $\gamma_2\sqrt[3]9 + \gamma_1\sqrt[3]3 + 1 = 0$, so
\begin{align*}0 &= (\gamma_2\sqrt[3]9 + \gamma_1\sqrt[3]3 + 1)^2\\[0.5em] &= (\gamma_1^2 + 2\gamma_2)\sqrt[3]9 + (2\gamma_1 + 3\gamma_2^2)\sqrt[3]3 + (1 + 6\gamma_1\gamma_2). \end{align*}
But we also have $(1 + 6\gamma_1\gamma_2)(\gamma_2\sqrt[3]9 + \gamma_1\sqrt[3]3 + 1) = 0$. Hence,
$$\sqrt[3]3 = \frac{\sqrt[3]9}{\sqrt[3]3} = \frac{(\gamma_1^2 + 2\gamma_2) - \gamma_2(1 + 6\gamma_1\gamma_2)}{\gamma_1(1 + 6\gamma_1\gamma_2) - (2\gamma_1 + 3\gamma_2^2)}.$$