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. 
 A: $$(x-\sqrt2)^3=2$$
$$\iff x^3-2+3x\cdot2=(3x^2+2)\sqrt2$$
Square both sides
A: The minimal polynomial of $\alpha=\sqrt{2}$ is $a^2-2$ and the minimal polynomial of $\beta=\sqrt[3]{2}$ is $b^3-2$. 
$\{1,a,b,b^2,ab,ab^2\}$ is a base of $\mathbb{A}=\mathbb{Q}[a,b]/(a^2-2,b^3-2)$, hence $\alpha+\beta$ is a root of a polynomial belonging to $\mathbb{Q}[x]$ with degree $\leq 6$. Indeed in $\mathbb{A}$ the terms $(a+b)^k$ decompose as follows:
$$\begin{array}{|c|c|c|c|c|c|c|}\hline  & 1 & a & b & b^2 & ab & ab^2\\
\hline 1 & 1 &&&&&\\
\hline (a+b) & 1 & 1 &&&&\\
\hline (a+b)^2 & 2 & & & 1 & 2 & \\
\hline (a+b)^3 & 2 & 2 & 6 &&& 3\\
\hline (a+b)^4 & 4 & 8 & 2 & 12 & 8 & \\
\hline (a+b)^5 & 40 & 4 & 20 & 2 & 10 & 20\\
\hline (a+b)^6 & 12 & 80 & 60  & 60 & 24 & 12\\
\hline\end{array}$$
hence by Gaussian elimination (we have seven rows and six columns)
$$ (a+b)^6 = 4+24(a+b)-12(a+b)^2+4(a+b)^3+6(a+b)^4 $$
and $\alpha+\beta$ is a root of $x^6-6x^4-4x^3+12x^2-24x-4$. This is the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$ since the involved matrix has rank $6$: the determinant of the matrix formed by the first six rows and columns is
$$ 924 = 3\cdot 2^2\cdot\det\left(\begin{smallmatrix}0&1&2&0\\2&0&0&1\\1&6&4&0\\10&1&5&10\end{smallmatrix}\right). $$
Your polynomial cannot vanish at $\sqrt{2}+\sqrt[3]{2}$ since the conjugates of this algebraic number are $\pm\sqrt{2}+\omega^k\sqrt[3]{2}$, hence the norm of $\sqrt{2}+\sqrt[3]{2}$ is $(-4)\color{red}{\neq(-23)}$.
A: 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.
