Let $\alpha$ be a root of $$ f(x)=x^{19}-7x^{16}+77x^{11}+63x^5-35x^3+14=0$$ Is $\mathbb{Q}(\alpha^8)=\mathbb{Q}(\alpha)$? Justify your answer.
Solution: Using Eisenstein's Criterion, with $p=7$, we show $f(x)$ is irreducible over $\mathbb{Q}$.
To show $\alpha^8$ is a root of $f(x)$, do I try to find $\alpha$ explicit?
Is there a trick to show $\mathbb{Q}(\alpha^8)=\mathbb{Q}(\alpha)$?
You have to show that $x$ and $x^8$ generate the same extension of $\mathbb{Q}$. Now, apriori, $\mathbb{Q}\subset \mathbb{Q}(x^8) \subset \mathbb{Q}(x)$. What you also know is that the degree of the extension $\mathbb{Q}(x)/\mathbb{Q}$ is $19$. Also, you know the formula for the degrees of the intermediate extension $$[\mathbb{Q}(x)\colon \mathbb{Q}] = [\mathbb{Q}(x)\colon \mathbb{Q}(x^8)]\cdot [\mathbb{Q}(x^8)\colon \mathbb{Q}]$$ Since the LHS is $19$, a prime, $[\mathbb{Q}(x^8)\colon \mathbb{Q}]$ can only be $1$ or $19$. It cannot be $1$, since $x^8\not\in \mathbb{Q}$. So it must be $19$, and therefore $\mathbb{Q}(x^8) = \mathbb{Q}(x)$.
Now, to provide an explicit equation for $x^8\colon = y$, we can use Groebner bases, see the WA calculation. Moreover, we can find an explicit expression of $x$ in terms of $y$. It is quite involved, see here.