Show that $\frac{-1+\sqrt{3}i}{2}$ is a cube root of 1? I know how to directly prove that complex number is one of the cube roots of 1. But the textbook solution also gives another solution as shown below,



However, I don't understand how it jumps to the conclusion that $\frac{-1+\sqrt{3}i}{2}$ is a root of the quadratic equation following the two equations at the top. Thank you!
Well maybe it is using Complex conjugate root theorem &amp; Vieta's formula to say that $\alpha+ \bar{\alpha}=-1$ and $\alpha \bar{\alpha}=1$. So $\alpha$ and $\bar \alpha$ are roots of $x^2-(-1)x+1$ or $x^2+x+1$. Hence $\alpha$ divides this which divides $x^3-1$ hence $\alpha$ is a root of $x^3-1$.

Note:
$\alpha = \frac{-1+\sqrt3i}2,\;\bar\alpha=\frac{-1-\sqrt3i}2$
take $x=\frac{-1+\sqrt{3}i}{2}$ show that $x^3=1$
$$x=\frac12(-1+\sqrt{3}i)\\x^3=\frac{1}{8}((-1)^3+(\sqrt{3}i)^3+3(\sqrt{3}i)(-1)(-1+\sqrt{3}i))=\\
\frac18 (-1-3\sqrt3i+3\sqrt3i-9i^2)=\\\frac18(-1-9i^2)=\\\frac18(-1+9)=\\1$$
other method
$$x^3-1=(x-1)(x^2-x+1)=0\\x=1\\x^2-x+1=0\\\to (x-\frac12)^2-(\frac14)+1=0\\(x-\frac12)^2=-(\frac34)\\(x-\frac12)=\pm\frac {\sqrt{-3}}{2}\\
(x-\frac12)=\pm\frac {\sqrt{3}i}{2}\\
x=\frac{1}{2}\pm\frac {\sqrt{3}i}{2}$$