<s>[INST] Are $a=3, b=7$ the only solutions to $\sqrt{3}+\sqrt[3]{7}=\sqrt{a}+\sqrt[3]{b}$ for $a,b \in \mathbb{Q}$? So I had a lesson about calculating surds e.g. $\sqrt{5+2\sqrt{6}}$, then the teacher wrote the steps like that:
For some $a,b\in\mathbb{Q}_{\ge 0}$
$$\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b} \\ 
5+2\sqrt{6}=a+b+2\sqrt{ab} \\
$$
which directly implies $\begin{cases} a+b=5 \\ ab=6\end{cases}$. The teacher didn't know the proof and I eventually found a proof of it:
Rewrite the equation, we get:
$$5-a-b+2\sqrt{6}=2\sqrt{ab}$$
Then let $2c=5-a-b$, where $c$ should be rational. Then we continue the equation:
$$ 2c+2\sqrt{6}=2\sqrt{ab} \\ c+\sqrt{6}=\sqrt{ab} \\ (c+\sqrt{6})^2=ab \\ c^2+6-ab=-2c\sqrt{6}$$
As $c^2+6-ab$ is rational, $-2c\sqrt{6}$ is also rational, which implies $c=0$, i.e. $a+b=5$. Substituting it back to the original equation, we get $ab=6$. Q.E.D.
He found the proof quite interesting and asked a question as follows:

For rational numbers $a,b$, is $(3,7)$ the only solution to the equation $\sqrt{a}+\sqrt[3]{b}=\sqrt{3}+\sqrt[3]{7}$?

I try to prove this but it is harder than it looks. The cubic root is one of the annoying part of proving this. What I am able to prove is that $\sqrt{3}+\sqrt[3]{7}$ is irrational. Other than that, I have no idea. To prove this, I think it may probably consist of algebraic fields, which I totally have no idea. I would like to know if there are any ways to solve this as this problem is quite interesting. If you guys have any ideas, please feel free to send it out. Thanks :)
 [/INST] By doing some calculations you can see that $\sqrt{3}+\sqrt[3]{7}$ is a root of the polynomial
$$
p(x) = x^6-9 x^4-14 x^3+27 x^2-126 x+22.
$$
Furthermore, $\sqrt{a}+\sqrt[3]{b}$ is a root of
$$
q_{a,b}(x) = x^6 -3 a x^4 -2 b x^3 +3 a^2 x^2 -6 a b x -a^3+b^2.
$$
The polynomial $p$ is irreducible. This means that it is the only monic degree 6 polynomial with rational coefficients of which $\sqrt{3}+\sqrt[3]{7}$ is a root. This means that if $a,b$ are rational such that $\sqrt{a}+\sqrt[3]{b}$ is equal to $\sqrt{3}+\sqrt[3]{7}$ that the polynomials $p$ and $q_{a,b}$ must be the same. Now you can look for example at the coefficients in front of the $x^3$ and $x^4$ terms to find that this implies that $a=3$ and $b=7$.
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