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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?

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If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then $$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$ This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1,\ \ x_1x_2 = 1. $$ So $x_1,x_2=(-1 \pm \sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.

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If $z_1$ and $z_2$ are roots of

$$(z-z_1)(z-z_2)=0$$

$$z^2-(z_1+z_2)z+z_1z_2=0$$

In particular, $z_1$ and $\bar{z_1}$ are roots of

$$z^2-2\Re(z_1)z+|z_1|^2=0$$

Hence, we can see that $\frac{-1+\sqrt3i}{2}$ is a root of $$x^2+x+1=0$$

Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$

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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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If $\frac{-1+i\sqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.

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