The degree of $\sqrt{2} + \sqrt[3]{5}$ over $\mathbb Q$

Here’s how you can show irreducibility of your polynomial, using the same method I used in the argument I gave in the reference that @BillDubuque mentioned:

First form $x^3-5$, the minimal polynomial for $\root3\of5$ over $\Bbb Q$. Now think of it as a polynomial over the principal ideal domain $\Bbb Z[\sqrt2\,]$. It happens that $5$ is still prime there, so Eisenstein still applies, and this is the minimal polynomial for $\root3\of5$ over the extension. Consequently $(x-\sqrt2)^3-5$ is still irreducible over $\Bbb Z[\sqrt2\,]$, and it’s the minimal polynomial for $\sqrt2+\root3\of5$. Let’s call this polynomial $f(x)$. It’s actually $x^3-3\sqrt2x^2+6x-5+2\sqrt2$. Now take the conjugate polynomial, call it $\bar f$, which you get by replacing $\sqrt2$ by $-\sqrt2$ in $f$ wherever it appears. Finally, multiply, and get $f\bar f$, which turns out to be exactly the sextic polynomial you calculated.

This clearly is a $\Bbb Q$-polynomial that has $\sqrt2+\root3\of5$ as a root. But it is also a $\Bbb Z[\sqrt2\,]$-polynomial, and we know its factorization into irreducibles there, namely $f\bar f$. And by uniqueness of factorization there, this is the only possible factorization with coefficients in $\Bbb Z[\sqrt2\,]$. But a $\Bbb Q$-factorization of your polynomial would also be a $\Bbb Z[\sqrt2\,]$-factorization, and we already have one such, and it’s unique. So there is no factorization of your polynomial over $\Bbb Z$ or $\Bbb Q$.

1. Let $V$ be the 6-dimensional vector space consisting of all numbers of the form $$a+b2^{1/2} + c5^{1/3} + d2^{1/2}5^{1/3} + e5^{2/3} + f2^{1/2}5^{2/3}$$ where $a,b,c,d,e,f\in \Bbb Q$

2. Let $w = \sqrt2+\sqrt[3]5$. The numbers $w^0, w^1, w^2, w^3, w^4, w^5$ are vectors from $V$. Since $V$ is 6-dimensional, these 6 vectors either span $V$ or else any them can be expressed as a linear combination of the others. Your undergraduate linear algebra class taught you how to check which it is.

3. If your $6$th-degree polynomial were reducible, it would have some factor $P$ of degree $d$ less than $6$; say $P(x) = x^d + Q(x)$. Since $P(w) = 0$, you have $$w^d = -Q(w)$$ where $Q$ has degree at most $d-1$. This expresses $x^d$ in terms of $x^0, x^1,\ldots, x^{d-1}$, contradicting your finding in step 2.

It's enough to show that $x^3-5$ is irreducible over $\mathbb{Q}(\sqrt{2})$. If it weren't, then it would have a root in $\mathbb{Q}(\sqrt{2})$. Since it has a unique real root, we must have $\sqrt[3]{5}\in \mathbb{Q}(\sqrt{2})$. So $(a+b\sqrt{2})^3=5$, and therefore $(a-b\sqrt{2})^3=5$, hence $a+b\sqrt{2}=a-b\sqrt{2}=\sqrt[3]{5}$, and so $\sqrt[3]{5}\in \mathbb{Q}$, contradiction.

$\bf{Added:}$ I am uncomfortable with my answer, since it assumes $\mathbb{Q}(\sqrt{2}+ \sqrt[3]{5}) = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5})$, which is the crux of the problem. If that is proved, then the degree has to be divisible by $2$, and $3$, so it is necessarily $6$. So let's just show that $\alpha = \sqrt{2}+ \sqrt[3]{5}$ generates both roots. Now, we have the equality $(\alpha - \sqrt{2})^3= 5$ from which we can obtain $\sqrt{2}$ as a rational fraction in $\alpha$. In detail: $$\alpha^3 - 3 \alpha^2 \sqrt{2} +3 \alpha \cdot 2 - 2 \sqrt{2} = 5$$ or $$\frac{\alpha^3 +6 \alpha - 5 }{3 \alpha^2 + 2} = \sqrt{2}$$ By the way, the equation that we get for $\alpha$ $$(x^3 + 6 x -5)^2 - 2(3x^2 + 2)^2=0$$ is the equation in the posting.