I have been told that $\mathbb{Q}(\sqrt[4]{3})$ as a vector space over $\mathbb{Q}$ has dimension $4$. If $\alpha = \sqrt[4]{3}$, then I am guessing a basis is $1, \alpha, \alpha^2, \alpha^3$. I can see that these elements span all of $\mathbb{Q}(\alpha)$, but I am not sure how to show that they are linearly independent.
Now, I think I can show that any two of the elements are linearly independent, but I know this doesn't imply that all four are linearly independent.
If the set $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly dependent over $\Bbb{Q}$, then there exist $x_0,x_1,x_2,x_3\in\Bbb{Q}$ not all zero such that $$x_0+x_1\alpha+x_2\alpha^2+x_3\alpha^3=0.$$ This means $\alpha$ is a root of the polynomial $f:=x_3X^3+x_2X^2+x_1X+x_0\in\Bbb{Q}[X]$. Because $\alpha$ is also a root of the polynomial $g:=X^4-3\in\Bbb{Q}[X]$, it is also a root of $\gcd(f,g)$. Now it suffices to show that $X^4-3$ is irreducible in $\Bbb{Q}[X]$ to reach a contradiction.