How can I get from one factored form to another?

Question

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?

Answer 1

$$\left(\frac{\sqrt{3}\pm i}{2}\right)^2 = \frac{3-1\pm 2\sqrt{3}i}{4} = \frac{1\pm\sqrt{3}i}{2} \implies \underbrace{\sqrt{\frac{1\pm \sqrt{3}i}{2}} = \frac{\sqrt{3}\pm i}{2}}_{\text{assume taking principal branch of } \sqrt{\;}} \\ {\Huge\Downarrow} \\ \left(x \pm \sqrt{\frac{1 + \sqrt{3}i}{2}}\right) \left(x \pm \sqrt{\frac{1 - \sqrt{3}i}{2}}\right)\\ \| \\ x^2 \pm \left(\frac{\sqrt{3} + i}{2} + \frac{\sqrt{3} - i}{2}\right) + \left(\frac{\sqrt{3} + i}{2}\right)\left(\frac{\sqrt{3} - i}{2}\right)\\ \| \\ x^2 \pm \sqrt{3}x + 1 $$

Answer 2

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 $$ \begin{split} x^6+1 &=\prod_{k=0}^{5}\left(x-e^{\frac{(2k+1)\pi}{6}i}\right)\\ &=\textstyle(x-\frac{\sqrt{3}}{2}-\frac{1}{2}i)(x-i)(x+\frac{\sqrt{3}}{2}-\frac{1}{2}i) (x+\frac{\sqrt{3}}{2}+\frac{1}{2}i)(x+i)(x-\frac{\sqrt{3}}{2}+\frac{1}{2}i) \end{split} $$