Let $c= \sqrt[4]{5}+\sqrt{5}$ and $d=\sqrt[4]{5}$ How can show that $\mathbb{Q}(c)=\mathbb{Q}(d)$?
I only have some idea about write $c=d+\sqrt{5}$ and do not know what I need to do for the next step.
Then, to find $[\mathbb{Q}(c):\mathbb{Q}]$ and the minimal polynomial of $a$ over $\mathbb{Q}$.
Even I do not know how to show for this question $[\mathbb{Q}(d):\mathbb{Q}]$ instead of $[\mathbb{Q}(c):\mathbb{Q}]$ since $\mathbb{Q}(c)=\mathbb{Q}(d)$. And, I know that the basis of $\mathbb{Q}(d)$ is $(1,\sqrt[4]{5},(\sqrt[4]{5})^2,(\sqrt[4]{5})^3)$. Hence, the degree of $[\mathbb{Q}(d):\mathbb{Q}]$ should be $4$.
To find the minimal polynomial of a over $\mathbb{Q}$, I end up with $x^4-x^2=0$ and in this case, it has not prime, therefore cannot apply Eisenstein's test.
You need to show containment both directions. Showing $\mathbb{Q}(c)\subseteq \mathbb{Q}(d)$ should be completely straightforward, since $c = d^2 + d$
For the other direction \begin{align*} c&= \sqrt[4]{5}+\sqrt{5} \\ c - \sqrt{5} &= \sqrt[4]{5} \\ c^2 - 2c \sqrt{5} + 5 &= \sqrt{5}\\ c^2 +5 &= (2c+1) \sqrt{5} \\ \frac{c^2+5}{2c+1} &= \sqrt{5} \end{align*} So we can conclude $$ c - \frac{c^2+5}{2c+1} = d $$ and hence $\mathbb{Q}(d)\subseteq \mathbb{Q}(c)$