Which of them are separable and which are normal?
That is the question and my respective answers follow below. I hope you have suggestions or if there are any mistakes I hope you help me fix them.
Normal: $\mathbb{Q}(\sqrt{7}, i)$
(i). irr$(1+\sqrt[3]{5}, \mathbb{Q}) = x^3 -3x^2+3x-6$ has a complex zero $\alpha = \dfrac{1}{2} \left[-\sqrt[3]{5} + 2 + \sqrt{-3 \sqrt[3]{25}}\right]$ (shown in #5 here: Consider the element $1+\sqrt[3]{5}$ of $\mathbb{R}.$) which is not in $\mathbb{Q}(1+\sqrt[3]{5}).$ Thus, $\mathbb{Q}(1+\sqrt[3]{5})$ is not normal over $\mathbb{Q}.$
(ii). $\mathbb{Q}(\sqrt{7}, i)$ is a splitting field for the polynomial $p(x) = (x^2 -7)(x^2+1)$ over $\mathbb{Q}.$ The roots $p(x)$ are $\pm\sqrt{7}, \pm i$ which all belong to $\mathbb{Q}(\sqrt{7}, i)$. Therefore, $\mathbb{Q}(\sqrt{7}, i)$ is a normal extension of $\mathbb{Q}.$
Separable: both
(i). irr$(1+\sqrt[3]{5}, \mathbb{Q}) = x^3 -3x^2+3x-6$ has distinct roots in its splitting field $\mathbb{Q}(1+\sqrt[3]{5}, i)$ implying that $x^3 -3x^2+3x-6$ is separable. Furthermore, this implies that every element of $\mathbb{Q}(1+\sqrt[3]{5})$ is separable over $\mathbb{Q}.$ Therefore, $\mathbb{Q}$: $\mathbb{Q}(1+\sqrt[3]{5})$ is separable.
(ii). The polynomial $p(x) = (x^2 -7)(x^2+1)$ has distinct roots. Thus, each of its zeroes is separable over $\mathbb{Q}$. Therefore, $\mathbb{Q}(\sqrt{7}, i)$ is separable.
Are there any comments? Are there any mistakes or can there be any improvements in my solutions? Thank you!