<s>[INST] How to prove $\sqrt{2} + \sqrt[3]{2}$ is an algebraic number I need to show that this is an algebraic number by showing that it is a solution to:
$x^{6} - 9x^{4} - 4x^{3} + 27x^{2} + 36x -23 = 0.$
I'm really struggling with this one, I tried squaring and cubing $\sqrt{2} + \sqrt[3]{2}$, but I only end up with more squares and cubes. 
 [/INST] Set
$\alpha = \sqrt 2 + \sqrt[3] 2; \tag 1$
then,
$\alpha - \sqrt 2 = \sqrt [3] 2; \tag 2$
$(\alpha - \sqrt 2)^3 = 2; \tag 3$
$\alpha^3 - 3 \sqrt 2 \alpha^2 + 6 \alpha - 2\sqrt 2 = 2; \tag 4$
$-\sqrt 2(2 + 3\alpha^2) = 2 - 6\alpha - \alpha^3; \tag 5$
$2(2 + 3\alpha^2)^2 = (2 - 6\alpha - \alpha^3)^2. \tag 6$
If all we want to do is show $\alpha$ algebraic over $\Bbb Q$, we can stop here; (6) is evidently a $6$ degree polynomial over $\Bbb Q$; 
also, a finer inspection of (6) shows $\alpha$ to be an algebraic integer, since the leading coefficient, of $\alpha^6$, is $1$; to see exactly what polynomial we get, however, we must expand out each side of (6):
$2(4 + 12\alpha^2 + 9\alpha^4) = 4 + 36\alpha^2 + \alpha^6 - 24\alpha - 4\alpha^3 + 12\alpha^4; \tag 7$
$8 + 24\alpha^2 + 18\alpha^4 = 4 + 36\alpha^2 + \alpha^6 - 24\alpha - 4\alpha^3 + 12\alpha^4; \tag 8$
finally,
$\alpha^6 - 6\alpha^4 - 4\alpha^3 + 12 \alpha^2 -24\alpha - 4 = 0, \tag 9$
a sixth-degree polynomial satisfied by $\alpha$, in agreement with that obtained by Jack D'urizio.
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