As the title says I have to show that these two extensions of the rational numbers are isomorphic: $$\mathbb{Q}(2^\frac1n; 3^\frac1m)=\mathbb{Q}(2^\frac1n3^\frac1m)$$ supposing that $\gcd(n,m)=1$
I really don't know how to even begin in the first place (as we never done a single exercise in class), the only thing I've managed to prove is that the two extension have the same dimension over $\mathbb{Q}$ and I feel very far from the actual solution so I'd prefer at least a sketch of the proof and not only a hint.
The inclusion $\mathbb{Q}(2^{1/n}3^{1/m})\subseteq\mathbb{Q}(2^{1/n},3^{1/m})$ should be clear. For the reverse inclusion, it suffices to establish $2^{1/n}$ and $3^{1/m}$ are elements of $\mathbb{Q}(2^{1/n}3^{1/m})$.
Hint #1: What happens when you take certain powers of $2^{1/n}3^{1/m}$?
Hint #2: Bezout's identity says if $\gcd(n,m)=1$ then there exist $a,b$ such that $an+bm=1$.
(You should be able to take $2^{1/n}3^{1/m}$ to a power until it is just one of $2$ or $3$ to a rational power, up to a rational multiple. And then you should be able to take said rational powers of $2$ or $3$ to another power so that it is $2^{1/n}$ or $3^{1/m}$ up to a rational multiple.)