## Verify that a given element is algebraic over the rationals

Show that $\sqrt{2} + \sqrt[3]{3}$ and $\sqrt{2}\sqrt[3]{3}$ are algebraic over $\mathbb{Q}$.

Certainly $(\sqrt{2}\sqrt[3]{3})^6 = 72$, so that $\sqrt{2}\sqrt[3]{3}$ is a root of $x^6 - 72$ and thus is algebraic over $\mathbb{Q}$.

Note that the conjugates of $\sqrt{2}$ are $\alpha_i = \pm \sqrt{2}$, and the conjugates of $\sqrt[3]{3}$ are $\beta_1 = \sqrt[3]{3}$, $\beta_2 = \sqrt[3]{3}(\frac{-1}{2} + \frac{\sqrt{-3}}{2})$, and $\beta_3 = \sqrt[3]{3}(\frac{-1}{2} - \frac{\sqrt{-3}}{2})$. Evidently, $\prod_i \prod_j (x - \alpha_i - \beta_j) = x^6-6x^4-6x^3+12x^2-36x+1$ has $\sqrt{2} + \sqrt[3]{3}$ as a root.