Say you want a polynomial with integer coefficients having a root with value $$v =\sum_{i=1}^k m_i\sqrt[q_i]{a}$$ where $k\ge 1$ and $\forall i: m_i, q_i \in \Bbb{Z}^+$, all the $q_i$ are greater than $1$ and distinct, and $a$ is an integer having no factors which are prime powers greater than $1$. (That is, $a$ is square-free, cube-free, and so forth.) For example, I might want a root to be $v=\sqrt[4]{7}+9\sqrt[6]{7}$.
It is always possible to construct a polynomial having $v$ as a root, by brute force of grouping terms to isolate one radical, raising to the appropriate power, and repeating until no radicals remain. (I will give an example below.) But this procedure leads to a polynomial which might be of higher degree than is necessary.
I conjecture that there always exists a polynomial of degree $\mbox{lcm}(q_1 \ldots q^k)$ (the least common multiple of the $\{q_i\}$) having $v$ as a solution.
Here is an example for which the conjecture holds: Say you want $x=\sqrt{a}+\sqrt[4]{a}$, then $$ \sqrt[4]{a}= x-\sqrt{a}\\ a = (x-\sqrt{a})^4 = x^4-4\sqrt{a}x^3+6ax^2-4a\sqrt{a}x+a^2 \\ (4x^3+4ax)\sqrt{a} = x^4+6ax^2+(a^2-a) \\ 16ax^2(x^2+a)^2=(x^4+6ax^2+(a^2-a))^2 \\ x^8-4ax^6+(6a^2-2a)x^4-(4a^3+12a^2)x^2+(a^2-a)^2 = 0 $$
This is a degree $8$ polynomial, but it factors into $$\left(x^4-2a x^2-4a x+(a^2-a)\right) \left(x^4-2a x^2+4a x+(a^2-a)\right)$$ and indeed $\sqrt{a}+\sqrt[4]{a}$ satisfies the quartic equation $$x^4-2a x^2-4a x+(a^2-a) =0. $$
On the other hand, to get a root of $\sqrt{a}+\sqrt[3]{a}$, $$ \sqrt[3]{a}= x-\sqrt{a}\\ a = (x-\sqrt{a})^3 = x^3-3\sqrt{a}x^2+3ax-a\sqrt{a} \\ (3x^2+a)\sqrt{a} = x^3+3ax-a \\ 9ax^4+6a^2x^2+a^3= x^6+6ax^4-2ax^3+9a^2x^2-6a^2x+a^2 \\ x^6-3ax^4-2ax^3+3a^2x^2-6a^2x+(a^2-a^3)= 0 $$ and that equation does not factor further (over the rationals) so the lowest degree is $6$, which again agrees with the conjecture.
The first really significant test of the conjecture would be with $q_1 = 4, q_6$. Here, the brute force method leads, after two iterations (raising to the $6$-th power, regrouping, and raising to the $4$-power), to a $24$-th degree equation, but that equation still contains terms involving $\sqrt{a}$. So the brute force method leads to an equation of degree $48$. The conjecture is that this factorizes, with at least one factor of degree $12$, and that (for instance) $\sqrt[4]{a}+\sqrt[6]{a}$ satisfies that degree $12$ equation.
Is this conjecture well known and established, or is it false or an open question?
Let $q = \mbox{lcm}\{q_1\ldots q_k\}$, $\alpha = \sqrt[q]a$ and $K = \Bbb Q[\alpha]$. $K$ is a field extension of $\Bbb Q$ of degree $q$.
Since every $\sqrt[q_i]a$ is a power of $\alpha$, there is a polynomial $P$ with rational coefficients such that $v=P(\alpha)$, that is, $v$ is an element of $\Bbb Q[\alpha]$, and so the degree of its minimal polynomial has to divide $q$.
Concretely, you can get a polynomial over $\Bbb Q$ of degree $q$ that annihilates $v$, for example by computing the characteristic polynomial of the multiplication-by-$v$ map, which is a linear endomorphism $K \to K$ of a $\Bbb Q$-vector space of dimension $q$.
You can also note that if $\zeta$ is a primitic $q$th root of unity, the conjugates of $\alpha$ are the $\zeta^k\alpha$ for $k=0 \ldots q-1$, and so if $v = P(\alpha)$ for some rational polynomial $P$, then its conjugates over $\Bbb Q$ are the $P(\zeta^k\alpha)$, and so the polynomial $\prod_{k=0}^{q-1} (X - P(\zeta^k \alpha))$ has rational coefficients and annihilates $v$.
Proving that those are the minimal polynomial of $v$ (so that they are not powers of a smaller polynomial) is slightly more difficult as you need to show that $\alpha \in \Bbb Q[v]$.