What are the roots of $x^4 + x^2 + 1$? I tried to find them for a quite a while but to no avail. I tried the substitution $y = x^2$, but the result is a mess. Is there any systematic way to get the roots for this quartic polynomial?
Any help or insight is deeply appreciated.
 A: The substitution $y=x^2$ is the right way.
$$y^2+y+1=0\iff y=\frac{-1\pm i\sqrt{3}}2$$
and 
$$x=\pm\sqrt{\frac{-1\pm i\sqrt{3}}2}.$$

There is a formula for the square roots of complex numbers,
$$\sqrt{a+ib}=\frac1{\sqrt 2}\left(\sqrt{\sqrt{a^2+b^2}+a}+i\text{ sgn}(b)\sqrt{\sqrt{a^2+b^2}-a}\right).$$
A: Hint: $x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2+1-x)(x^2+1+x)$
A: Notice that:
$$x^4+x^2+1=(x^2 +1)^2-x^2$$
Differences of squares identity should work reliably here. The remaining should be easy work.
Clearly, there are 4 roots to answer your problem. Straightforward huh?
A: Two possible approaches:


*

*solve $y^2+y+1=0$ and then $x^2=y$,

*observe that $(x^4+x^2+1)(x^2-1)=x^6-1$.

A: Let $y := x^2$. The solution set of $y^2 + y + 1 = 0$ is
$$y \in \left\{ \frac{-1 \pm i\sqrt{3}}2 \right\} = \left\{ \exp \left( i \left( \pm\frac{2 \pi}{3} + 2 k \pi \right) \right) \mid k \in \mathbb Z \right\}$$
Taking the square root, we obtain the solution set of the original quartic equation
$$x \in \left\{ \exp \left( i \left( \pm\frac{\pi}{3} + k \pi \right) \right) \mid k \in \mathbb Z \right\} \color{white}{= \left\{ \exp \left( i \frac{\pi}{3} \right), \exp \left( i \frac{2\pi}{3} \right), \exp \left( i \frac{4\pi}{3} \right), \exp \left( i \frac{5\pi}{3} \right) \right\}}$$
