Are the fields $\mathbb{Q}(\sqrt[7]{16}+3 \sqrt[7]{8})$ and $\mathbb{Q}(\sqrt[7]{16})$ equal? I have trouble with these field extensions. Is field  $\mathbb{Q}(\sqrt[7]{16}+3 \sqrt[7]{8})$ equal to field $\mathbb{Q}(\sqrt[7]{16})$?
We can $\sqrt[7]{16}+3 \sqrt[7]{8}$ express as $(\sqrt[7]{2}+3)(\sqrt[7]2)^3$.
 A: Yes, they are the same. Observe that
$$(\sqrt[7]{16})^6=(2^{4/7})^6=2^{24/7}=8\cdot\sqrt[7]{8}$$
so that $\mathbb{Q}(\sqrt[7]{16}+3\sqrt[7]{8})\subseteq\mathbb{Q}(\sqrt[7]{16})$. We have $[\mathbb{Q}(\sqrt[7]{16}):\mathbb{Q}]=7$, and clearly $\sqrt[7]{16}+3\sqrt[7]{8}\notin\mathbb{Q}$, so 
$$7=[\mathbb{Q}(\sqrt[7]{16}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[7]{16}):\mathbb{Q}\sqrt[7]{16}+3\sqrt[7]{8})]\underbrace{[\mathbb{Q}(\sqrt[7]{16}+3\sqrt[7]{8}):\mathbb{Q}]}_{\large >\, 1}$$
Since $7$ is prime, we must have $[\mathbb{Q}(\sqrt[7]{16}):\mathbb{Q}(\sqrt[7]{16}+3\sqrt[7]{8})]=1$ and $[\mathbb{Q}(\sqrt[7]{16}+3\sqrt[7]{8}):\mathbb{Q}]=7$. Thus, we must have  $\mathbb{Q}(\sqrt[7]{16}+3\sqrt[7]{8})=\mathbb{Q}(\sqrt[7]{16})$.
Alternatively, you can use a computer algebra system or some tedious computations to find that the minimal polynomial for $\sqrt[7]{16}+3\sqrt[7]{8}$ over $\mathbb{Q}$ is
$$-17512 - 1512 x + 504 x^3 - 42 x^5 + x^7 $$
A: $$(\sqrt[7]{16})^6=8\sqrt[7]8$$
This shows that $\Bbb Q[\sqrt[7]{16}]\supseteq\Bbb Q[\sqrt[7]{16}+3\sqrt[7]8]$.
Now, consider the polynomials $x^4+3x^3$ and $9x^6+2x+6$. They are coprime, so we can apply the Bezout's identity (note that $\Bbb Q[x]$ is an Euclidean domain) to guarantee that there exist some polynomials $P$ and $Q$ such that
$$P(x)(x^4+2x^3)+Q(x)(9x^6+2x+6)=x^4$$
Let $a=\sqrt[7]2$. Then we find that
$$(\sqrt[7]{16}+3\sqrt[7]8)^2=(a^4+3a^3)^2=9a^6+2a+6$$
Therefore,
$$P(a)(a^4+3a^3)+Q(a)(a^4+3a^3)^2=a^4$$
This shows that $\Bbb Q[\sqrt[7]{16}]\subseteq\Bbb Q[\sqrt[7]{16}+3\sqrt[7]8]$
