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