<s>[INST] How to show that $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ is algebraic? How to show that $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$  is algebraic?
I know how to prove x = $\sqrt[3]{3} + \sqrt{2}$ and  x = $\sqrt{3} + \sqrt{2}$ , but the root doesn't disappear from $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ .
I want to make an algebraic equation that has this as an solution.
 [/INST] You can show that the polynomial of degree $2\cdot 2 \cdot 3 = 12$
$$\prod_{\epsilon^2 = \epsilon'^2 = \omega^3=1} ( x -\epsilon \sqrt{2} -\epsilon' \sqrt{3} - \omega \sqrt[3]{3})$$
has integral coefficients and your number as a root.
In general you cook up a polynomial formed from all possible sums of roots in a similar way ( here use roots of $x^2-2$, $x^2-3$, $x^3-3$ ).
More hints: the above polynomial equals
$$\prod ((x\pm \sqrt{2}\pm \sqrt{3})^3 - 3)$$
</s>