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. 

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 $(-4)\color{red}{\neq(-23)}$.
