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.
It is, because it's iterated application of addition, multiplication and radicals. These operations, together with polynomial root solving in general (including simple radicals) generate the set of algebraic numbers, which is closed for these operations. That alone is a proof that your number is a root of some polynomial (as it is an algebraic number).
Unraveling the number is actually a very easy construction of the polynomial you're looking for. Just perform algebraic operations on your root until you get an integer.
$$x=\sqrt[3]{2}+\sqrt{1+\sqrt{2}}$$ get rid of the cubic root $$(x-\sqrt{1+\sqrt{2}})^3=2$$ expand $$x^3-3x^2 \sqrt{1-\sqrt{2}}+3x(1-\sqrt{2})-\sqrt{1-\sqrt{2}}(1-\sqrt{2})=2$$ rearrange to kill the double roots in the next step $$x^3+3x(1-\sqrt{2})-2=(3x^2+1-\sqrt{2})\sqrt{1-\sqrt{2}}$$ square again $$(x^3+3x(1-\sqrt{2})-2)^2=(3x^2+1-\sqrt{2})^2(1-\sqrt{2})$$ We're almost there, the only thing left is the $\sqrt{2}$. Expand all terms (I'm not doing that, it's gnarly), group those with $\sqrt{2}$ on one side, and square again. You will get a polynomial of order $12$ with one of the roots equal to your initial number.
You needn't explicitly show a polynomial, which would be very cumbersome for, say, $$ \sqrt[5]{3-\sqrt[3]{41}}+\sqrt[91]{101-\sqrt{2}} $$ You can, instead, use a more general result: if you add an algebraic number to a field consisting of algebraic numbers, the resulting field also consists of algebraic numbers.
An algebraic number $r$ is a complex number such that there exists a polynomial with rational coefficients (or, equivalently, integer coefficients) having $r$ as root.
This is based on some general lemmas.
Lemma 1. If $K$ is a finite extension of the field $F$, then every element of $K$ is algebraic over $F$.
Proof. Let $b\in K$ and assume the dimension of $K$ as vector space over $F$ is $n$ (the dimension is finite by assumption). Then the set $\{1,b,b^2,\dots,b^n\}$ is linearly dependent over $F$, which means there are $a_0,a_1,\dots,a_n\in F$, not all zero, such that $a_0+a_1b+\dots+a_nb^n=0$. Therefore $b$ is a root of the polynomial $a_0+a_1X+\dots+a_nX^n\in F[X]$.
Lemma 2. If $K$ is a finite extension of the field $F$ and $L$ is a finite extension of the field $K$, then $L$ is a finite extension of the field $F$.
Sketch of proof. Prove that, if $\{b_1,\dots,b_n\}$ is a basis for $K$ over $F$ and $\{c_1,\dots,c_m\}$ is a basis for $L$ over $K$, then $\{b_ic_j:1\le i\le n,1\le j\le m\}$ is a basis for $L$ over $F$.
Lemma 3. *If $K$ is an extension of the field $F$ and $b\in K$ is algebraic over the field $F$, then the field $F(b)$ (the minimum subfield of $K$ containing $F$ and $b$) is a finite extension of $F$.
Sketch of proof. If $a_0+a_1X+\dots+a_nX^n\in F[X]$ is the minimal polynomial of $b$ over $F$ (any non zero polynomial in $F[X]$ having $b$ as root would suffice), then $\{1,b,\dots,b^n\}$ is a spanning set for $F(b)$ as a vector space over $F$.
Now you can observe that $\sqrt[3]{2}+\sqrt{1+\sqrt{2}}\in\mathbb{Q}(\sqrt{1+\sqrt{2}})(\sqrt[3]{2})$. Since $\sqrt[3]{2}$ is obviously algebraic over $\mathbb{Q}$, it is also algebraic over $\mathbb{Q}(\sqrt{1+\sqrt{2}})$. Moreover $\sqrt{1+\sqrt{2}}$ is algebraic over $\mathbb{Q}(\sqrt{2})$, which in turn is algebraic over $\mathbb{Q}$.
Repeated applications of the lemmas show that $\mathbb{Q}(\sqrt{1+\sqrt{2}})(\sqrt[3]{2})$ is a finite extension of $\mathbb{Q}$, so any of its elements is algebraic over $\mathbb{Q}$.
$$\mathbb{Q}[\alpha,\beta]/(\alpha^3-2,\beta^4-2\beta^2-1)$$ 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: $$x^{12}-6x^{10}+8x^9 +9 x^8+28 x^6-144x^5 + 63 x^4+96x^3-78 x^2-168 x-41. $$
