In exercise 1A Q2, the question wants us to show that $\dfrac{-1+\sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said: Note that: $$(a+bi)+(a-bi)=2a$$ , and $$(a+bi)(a-bi)=a^2+b^2$$
It follows that $\dfrac{-1+\sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$\frac{-1+\sqrt{3}i}{2}+\frac{-1-\sqrt{3}i}{2}=-1$$ and $$\frac{-1+\sqrt{3}i}{2}\frac{-1-\sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?