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.

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