splitting field of $X^{4} -42$ over $\mathbb{Q}$

Question

We have given a intermediate body $K = \mathbb{Q} ( 42^{1/4},i)$ in $\mathbb{C}|\mathbb{Q}$

The question is now:

1. Are all complex zero points of $X^4-42$ in $K$ and/or in $\mathbb{Q} ( 42^{1/4},i)$?

   2. What is $a,b,c,d \in \mathbb{Q}$ with $\frac{1}{42^{1/4}-1} = a+b \cdot42^{1/4}+c \cdot42 +d \cdot 42^{1/2} \cdot 42^{1/4}$

Thanks in advance

Answer 1

$f(x)=x^4-42=(x-\alpha)(x+\alpha)(x+i\alpha)(x-i\alpha)\in \mathbb Q(\alpha,i)[x] $, so $\mathbb Q(\alpha,i)$ is a splitting field of $X^4−42$ over $\mathbb Q$, where $\alpha= \ ^{4}\sqrt{42} $.