Is $\mathbb{Q}(2^{1/6})\cong\mathbb{Q}(\sqrt2)(2^{1/6})$ true?

Question

Solving a problem I have seen that

$m_{(2^{1/6},\mathbb{Q}(\sqrt 2))}=x^3-\sqrt2$ , $[\mathbb{Q}(2^{1/6}):\mathbb{Q}(\sqrt2)]=3$

Because $m_{(2^{1/6},\mathbb{Q}(\sqrt 2))}=x^3-\sqrt2$, $$\mathbb{Q}(\sqrt2)[x]/(x^3-\sqrt2)\cong\mathbb{Q}(\sqrt2)[2^{1/6}]\cong\mathbb{Q}(\sqrt2)(2^{1/6})$$

Does this mean that $\mathbb{Q}(2^{1/6})\cong\mathbb{Q}(\sqrt2)(2^{1/6})$?

Answer 1

$\mathbb{Q}(\sqrt2)(2^{1/6})= \mathbb{Q}(\sqrt2,2^{1/6}) = \mathbb{Q}(2^{1/6})$ because $\sqrt2=(2^{1/6})^3 \in \mathbb{Q}(2^{1/6})$.