<s>[INST] Need help on understanding the solution (Linear Algebra done right) 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?
 [/INST] The solution is trying to form a quadratic equation one of whose roots is $\frac{-1+i\sqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $\frac{-1-i\sqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots. 
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$\frac{-1+\sqrt{3}i}{2}+\frac{-1-\sqrt{3}i}{2}=-1=-b/a, \frac{-1+\sqrt{3}i}{2}\cdot\frac{-1-\sqrt{3}i}{2}=1=c/a\implies a=c=b=1$
The required quadratic equation is $x^2+x+1=0\implies x^2=-x-1\implies x^3=-x^2-x=1$
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