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up vote 2 down vote favorite 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? linear-algebra share | cite | improve this question ...