Let $a,b \in \mathbb{Z}$ be integers such that $\sqrt{a} \notin \mathbb{Z}$ and $ \sqrt[3]{b} \notin \mathbb{Z}$ (the number $a$ is allowed to be negative). I need to prove that
$$ \mathbb{Q}(\sqrt{a}+\sqrt[3]{b}) = \mathbb{Q}(\sqrt{a}, \sqrt[3]{b})$$
but I cannot use any Galois theory to do so, as we have not gotten to it yet in my course.
I was given a hint: "let $G$ and $H$ denote the left-hand side and the right-hand side, respectively; analyze possibilities for the dimension $[H:G]$ and use thee fact that $G(\sqrt{a}) = G(\sqrt[3]{b}) = H$".
However, instead, I approached it similarly to Paramanand Singh's answer to this question. But, I am not sure such an approach is correct in this situation.
Anyway, this is what I did, according to Paramanand's method:
It is obvious that $\mathbb{Q}(\sqrt{a} + \sqrt[3]{b}) \subseteq \mathbb{Q}(\sqrt{a}, \sqrt[3]{b})$, since $\sqrt{a} + \sqrt[3]{b}$ is a linear combination of $\sqrt{a}$ and $\sqrt[3]{b}$
Now, for the other side, let $c = \sqrt{a}+\sqrt[3]{b}$.
WTS: $e=\sqrt{a}$, $f = \sqrt[3]{b}$ are rational functions of $c$.
$$ (c-e)^{3} = b \, \to \, c^{3}-3c^{2}e + 3ac - ae = b \, \to \, 3c^{2}e + ae = -b + c^{3}+3ac \, \to \, e = \frac{c^{3} + 3ac - b}{3c^{2} + a}\, \to \, \sqrt{a} = \frac{c^{3} + 3ac -b }{3c^{2}+a} $$
So that $\sqrt{a}$ is a rational function of $c$.
Also, since $c = e + f = \sqrt{a} + \sqrt[3]{b}$, we have that $\sqrt[3]{b} = c - \sqrt{a}$, so that $\sqrt[3]{b}$ is also a rational function of $c$.
Therefore $\mathbb{Q}(\sqrt{a}, \sqrt[3]{b}) \subseteq \mathbb{Q}(\sqrt{a} + \sqrt[3]{b})$
However, something about this is not sitting right with me. Is this, in fact, the correct way to prove it? Or should I do it the way my professor suggested? I would also like some help doing it that way.
I guess for that way, we could still start out by stating that "It is obvious that $\mathbb{Q}(\sqrt{a} + \sqrt[3]{b}) \subseteq \mathbb{Q}(\sqrt{a}, \sqrt[3]{b})$, since $\sqrt{a} + \sqrt[3]{b}$ is a linear combination of $\sqrt{a}$ and $\sqrt[3]{b}$" (if this is, in fact, true). Then, $\mathbb{Q}(\sqrt{a}, \sqrt[3])$ has degree $6$, right?
So, I would need to show that $\mathbb{Q}(\sqrt{a} + \sqrt[3]{b})$ also has degree $6$, but I still don't see what this has to do with the hint...
If I let $G = \mathbb{Q}(\sqrt{a} + \sqrt[3]{b})$ and $H = \mathbb{Q}(\sqrt{a}, \sqrt[3]{b})$, then of course $G(\sqrt{a}) = \mathbb{Q}(\sqrt{a} + \sqrt[3]{b})(\sqrt{a}) = \mathbb{Q}(\sqrt{a} + \sqrt[3]{b})(\sqrt[3]{b})$, but I still don't see how this helps!
Could someone please explain this to me? I would be most appreciative! Thank you!
Hint about your instructor's hint: The hint suggests that you look at $[H:G]$ by considering $[G[\sqrt{a}]:G]$ and $[G[\sqrt[3]{b}]:G]$ which both equal $[H:G]$. Now since $\sqrt{a}\not\in\Bbb{Z}$ and $\sqrt[3]{b}\not\in\Bbb{Z}$, $x^2-a$ and $x^3-b$ are irreducible over $\Bbb{Z}$. Looking at each of those polynomials over $G$ should give you a possibility for $[H:G]$ and considering them together will give the answer.