Show that $\mathbb{F}(w)$ is the splitting field of $x^4+x^2+1$, where $w = (-1+i\sqrt{3})/2$
All I need to show is that the roots of $x^4+x^2+1$ are in $\mathbb{F}(w)$?
The roots are:
$$\pm \sqrt[3]{-1}, \pm (-1)^{\frac{2}{3}}$$
I know I need to add combinations of $\pm(-1+i\sqrt{3})/2$ and $\pm(-1+i\sqrt{3})/2$ times these combinations, but which field is $\mathbb{F}$? Shouldn't I need to know it so I can combine with the combinations of the root? Or maybe I can form all the roots with just adding and multiplying this $\pm w$?
Let's see:
$$w^2 = \frac{1-2i\sqrt{3} -3}{4} = \frac{-2-2i\sqrt{3}}{4} = \frac{-1-i\sqrt{3}}{2} = -\left(\frac{1+i\sqrt{3}}{2}\right)\implies \pm w^2 = \mp \sqrt[3]{-1}$$
The root $(-1)^{\frac{2}{3}}$ is simply $w$, right?
So is this it?
Note that $$x^4+x^2+1=\frac{x^6-1}{x^2-1}$$ so that its zeroes are the sixth roots of unity, other than $\pm1$. If $w=\frac12(-1+i\sqrt3)$ then the zeroes of the polynomial are $w$, $-w$, $w^2$ and $-w^2$, so it splits over $\Bbb Q(w)$.