Find a polynomial with integer coefficients

We have $$ a=\sqrt2+\sqrt[3]2\\ a-\sqrt2=\sqrt[3]2\\ (a-\sqrt2)^3=2\\ a^3-3\sqrt2a^2+6a-2\sqrt2=2\\ a^3+6a-2=\sqrt2(3a^2+2)\\ (a^3+6a-2)^2=2(3a^2+2)^2 $$ which has all integer coefficients, once you expand the brackets.

There is a simple way to do this. Consider a matrix $M_1$ whose eigenvalues are $\pm \sqrt{2},$ and a matrix $M_2,$ whose eigenvalues are the roots of $x^3-2$ - this is easy to do by the companion matrix construction. Now, consider the matrix $M_1 \otimes I_3 + I_2 \otimes M_2 ,$ where $I_k$ is the $k\times k$ identity matrix. Its characteristic polynomial is what you are looking for.

Let a row vector, $r$, represent a real number by $$ r\, \begin{bmatrix} 1\\2^{1/6}\\2^{2/6}\\2^{3/6}\\2^{4/6}\\2^{5/6} \end{bmatrix} $$ multiplication of the row vector $r$ by $2^{1/3}+2^{1/2}$ can represented as $rM$ where $$ M= \begin{bmatrix} 0&0&1&1&0&0\\ 0&0&0&1&1&0\\ 0&0&0&0&1&1\\ 2&0&0&0&0&1\\ 2&2&0&0&0&0\\ 0&2&2&0&0&0\\ \end{bmatrix} $$ The top row of $M^k$ gives the representation of $\left(2^{1/3}+2^{1/2}\right)^k$. If $P$ is the characteristic polynomial of $M$, then $P(M)=0$. Thus, we also have $P\!\left(2^{1/3}+2^{1/2}\right)=0$, since that is represented by the top row of $P(M)$.

The characteristic polynomial of $M$ is $$ P(x)=x^6-6x^4-4x^3+12x^2-24x-4 $$ Therefore, $$ P\!\left(2^{1/3}+2^{1/2}\right)=0 $$