If $f(x)$ is irreducible, is $f(x^k)$ irreducible? Let $f(x)\in\mathbb{Z}[x]$ be an irreducible polynomial of degree $\ge 2$. Is it true that $f(x^k)$ is irreducible for $k\ge 2$? If not true, under what hypothesis, we can gurantee positive answer?

For $\alpha=\sqrt[6]{\sqrt{2}+\sqrt{3}}$, I saw that it satisfies a polynomial over $\mathbb{Q}$ given by $x^{24}-10x^{12}+1$. I wanted to check whether this polynomial is irreducible, and I came to above general question.
 A: Here is a sufficient condition:
$f(x)$ is irreducible over $\mathbb{Q}$ and in some field $K$ (the splitting field of $f(x)$) we have
$f(x) = (x-\alpha_1)\cdot \ldots \cdot (x-\alpha_n)$, and moreover each $x^k - \alpha_i$, $1\le i \le n$, is irreducible over $K$.
This applies in particular to your polynomial, see this answer.
$\bf{Added:}$ The point is that  over  some extension $K$ the polynomial $f(x^k)$ factors into irreducibles that are polynomials in $x^k$.
$\bf{Added:}$ Let $\alpha$ be any root of $f(x)$. We have $f(x^k)$ irreducible if and only if the degree of $\sqrt[k]{\alpha}$ over $\mathbb{Q}$ is $k n$. This is equivalent to: the degree of $\sqrt[k]{\alpha}$ over $\mathbb{Q}(\alpha)$ is $k$.
A: If $k\ne2$ then
$f(x)=x^2-2^k$ is irreducible but
$f(x^k)=(x^2)^k-2^k$ is reducible
A: $$f(x) = x^2 + 1$$ is irreducible over $\mathbb Q$, but $$f(x^3) = x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$$ obviously factors.
A: Another sufficient condition is given in Problems from the Book by Titu Andreescu (Example 9, page 494):

Let $f(x)$ be a monic polynomial with integer coefficients and let $p$ be a prime number. If $f(x)$ is irreducible in $\mathbb{Z}[x]$ and $\sqrt[p]{(-1)^{\deg f}f(0)}$ is irrational, then $f(x^p)$ is also irreducible in $\mathbb{Z}[x]$.

