$[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}):\Bbb Q(\sqrt[p_1]{q_1})] = p_2$ My original question is

$[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3}):\Bbb Q]=p_1p_2p_3$ ? where $p_1,p_2,p_3,q_1,q_2,q_3$ are all distinct prime numbers.

But I would rather change the question to $[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}):\Bbb Q(\sqrt[p_1]{q_1})] = p_2$ i.e. $x^{p_2}-q_2$ is irreducible over $\Bbb Q(\sqrt[p_1]{q_1})$. Any hints or ideas?
Edit: I don't know whether the above statement is true or not. But I guess it's true but I have no idea how to prove it.
 A: First, I don't think this question deserves the downvotes, as you make a reasonable attempt to replace your problem with a simpler one to understand it, and I understand what you're asking.
Here is a hint to help you think about your simpler problem, and I believe it will also get you to an answer to your original question.  Consider the field lattice diagram consisting of $ F = \mathbb{Q} $, the two extensions $ E_1 = \mathbb{Q}(\sqrt[p_1]{q_1}) $ and $ E_2 = \mathbb{Q}(\sqrt[p_2]{q_2} $), and the full field $ K = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}) $.  Then (assuming they are non-trivial) these extensions $ E_1/F $ and $ E_2/F $ have index $ p_1 $ and $ p_2 $.  Now here is the punchline:
$$
[K : F] = [K : E_1][E_1 : F] = [K:E_2][E_2:F]
$$
i.e. $ [K:F] $ is divisible by both $ p_1 $ and $ p_2 $.  This means $ [K : F] $ is divisible by $ p_1 p_2 $ (!).  I believe you can take it from here to conclude that $ K = \mathbb{Q}(\sqrt[p_1]{q_1})(\sqrt[p_2]{q_2}) $ is a degree-$p_2$ extension of $ E_1 = \mathbb{Q}(\sqrt[p_1]{q_1}) $.
