What is the factorization of $p(x) = x^3-10x+4$ over $ \mathbb Q(\sqrt 2, \sqrt[3] 2)?$
Since $p(x)$ is of degree $3$, it is irreducible over any field extension of $\mathbb Q$ if and only if no roots of $p(x)$ lie in that field. So, let us fix $\alpha$ to be a root of $p(x)$.
We first notice that $p(x)$ is irreducible over $\mathbb Q$, by the rational root test: none of the elements $1,2,4$ are roots of $p(x)$. Thus, $[\mathbb Q(\alpha):\mathbb Q] = 3.$
We see that $p(x)$ is also irreducible over $\mathbb Q(\sqrt 2)$. Indeed, if $\alpha \in \mathbb Q(\sqrt 2)$, then $3 = [\mathbb Q(\alpha) : \mathbb Q]$ would have to divide $[\mathbb Q(\sqrt 2) : \mathbb Q] = 2$, which is not the case. As a consequence, we get $[\mathbb Q(\sqrt 2, \alpha) : \mathbb Q(\sqrt 2) ] =3.$
We also note that $[\mathbb Q(\sqrt 2, \sqrt[3] 2):\mathbb Q(\sqrt 2)] = 3$. This follows from the fact that $[\mathbb Q(\sqrt 2, \sqrt[3]2) : \mathbb Q ] = 6.$ Now, using the Field Tower Theorem, we get
$$3 = [\mathbb Q(\sqrt[3] 2, \sqrt 2): \mathbb Q(\sqrt 2)] = [\mathbb Q(\sqrt 2, \sqrt[3] 2): \mathbb Q(\sqrt 2, \alpha)]\cdot [\mathbb Q(\sqrt 2, \alpha) : \mathbb Q(\sqrt 2)] = 3 [\mathbb Q(\sqrt 2, \sqrt[3] 2): \mathbb Q(\sqrt 2, \alpha)].$$
Therefore, $[\mathbb Q(\sqrt 2, \sqrt[3] 2): \mathbb Q(\sqrt 2, \alpha)] = 1$ shows that $\alpha \in \mathbb Q(\sqrt[3] 2, \sqrt 2)$. Hence, $p(x)$ is reducible over $\mathbb Q(\sqrt[3] 2, \sqrt 2)$.
However, I have no idea how to find the factorization of $p(x)$ over this field. This mainly comes from the fact that I dont have simple enough expression for the roots of $p(x)$, so I dunno how to manipulate them into elements that belong to the field $\mathbb Q(\sqrt[3] 2, \sqrt 2)$. Can somebody give me an idea how to perform that? Or point out if theres any flaws in my arguments, so that in fact $p(x)$ is irreducible over this field too.
There is a way to solve cubic equation in the form $x^3 + bx + c = 0$, you will do a sub $x = u+\frac{a}{u}$ find $a$ that eliminate the $u$ and $1/u$, in our case $$(u+au^{-1})^3 + b(u+au^{-1}) + c = 0$$ $$u^3 + 3au+ 3a^2u^{-1} + a^3u^{-3} + bu + abu^{-1} + c=0$$ We want $a$ such that $3a= -b \Rightarrow a = \frac{-b}{3}$ Doing that you will end up with $$u^3 + a^3 u^{-3} + c = 0$$ Which is $$(u^3)^2 + c u^3 + a^3 =0 $$ Solve as quadratic.