<s>[INST] How can I get from one factored form to another? This came from trying to factor $x^6 + 1$. We know that 
$x^4 - x^2 + 1 =  (x^4 +2 x^2 + 1) -3x^2 = (x^2+1)^2 - 3x^2 = (x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$
I expected that if I factored the $x^4 - x^2 + 1$ completely into $4$ factors and then regrouped them $2$ by $2$, I could arrive at $(x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$
$$\left(x+ \sqrt {\dfrac {1 + \sqrt 3 i}{2}}\right) \left(x+ \sqrt {\dfrac {1 - \sqrt 3 i}{2}}\right) \left(x- \sqrt {\dfrac {1 + \sqrt 3 i}{2}}\right) \left(x- \sqrt {\dfrac {1 - \sqrt 3 i}{2}}\right)$$
But that didn't happen. No matter how I group them into quadratics, I always end up with a complex coefficient. How can I get $(x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$ from the $4$ factors?
 [/INST] Solve the equation
$$
x^6=-1
$$
using polar coordinates: $-1=e^{\pi i}$, so
$$
x_k=e^{\frac{\pi+2\pi k}{6}i}=e^{\frac{(2k+1)\pi}{6}i},
$$
for $k\in\{0,1,2,3,4,5\}$. Therefore 
$$
\begin{split}
x^6+1 &=\prod_{k=0}^{5}\left(x-e^{\frac{(2k+1)\pi}{6}i}\right)\\
&=\textstyle(x-\frac{\sqrt{3}}{2}-\frac{1}{2}i)(x-i)(x+\frac{\sqrt{3}}{2}-\frac{1}{2}i)
(x+\frac{\sqrt{3}}{2}+\frac{1}{2}i)(x+i)(x-\frac{\sqrt{3}}{2}+\frac{1}{2}i)
\end{split}
$$
</s>