Prove or disprove that $\sqrt[3]{2}+\sqrt{1+\sqrt2}$ is a root of a polynomial Prove or disprove that there is a polynomial with integer coefficients such that the number $\sqrt[3]{2}+\sqrt{1+\sqrt2}$  is a root.
My work so far:
Let $P(x)=x^3-2$. Then $\sqrt[3]{2}$ is a root of $P(x)$
Let $Q(x)=x^4-2x^2-1$. Then $\sqrt{1+\sqrt2}$ is a root of $Q(x)$.
$\left(x=\sqrt{1+\sqrt2}\Rightarrow x^2=1+\sqrt2\Rightarrow x^2-1=\sqrt2 \Rightarrow x^4-2x^2+1=2 \Rightarrow x^4-2x^2-1=0\right)$
I need help here.

$$\mathbb{Q}[\alpha,\beta]/(\alpha^3-2,\beta^4-2\beta^2-1)$$
is a vector space over $\mathbb{Q}$ with dimension $12$: a base is given by $\alpha^n \beta^m$ for $0\leq n\leq 2$ and $0\leq m \leq 3$. If follows that if we represent $(\alpha+\beta)^k$ for $k=0,1,\ldots,11,12$ with respect to such a base, we get $13$ vectors in a vector space with dimension $12$, hence we may find a linear combination of them that equals zero, hence a polynomial with coefficients in $\mathbb{Q}$ and degree $\leq 12$ that vanishes at $\alpha+\beta$. Namely, such a polynomial is:
$$x^{12}-6x^{10}+8x^9 +9 x^8+28 x^6-144x^5 + 63 x^4+96x^3-78 x^2-168 x-41. $$
