Solving roots of unity

Question

The three cube roots of unity are

$\omega$, $\omega^2$ and 1, where $$\omega = \frac{-1+\iota\sqrt{3}}{2}, \qquad\omega^2 = \frac{-1-\iota\sqrt{3}}{2}$$

Evaluating the equation

$$x^3 - y^3 = (x-y)(x-\omega y)(x - \omega^2y)$$ My attempt : Solving R.H.S

$$ (x-y)(x^2=\omega^2xy-\omega xy+\omega^3y^2) $$ $$(x-y)(x^2-\omega(\omega xy-xy) + y^2)\space \because \space \omega^3 = 1$$

I'm kinda confused what to do with now. any help ?

Please notices that $\omega^2+\omega+1=0$. – Riemann Aug 11 '18 at 11:26 — Riemann, Aug 11 '18 at 11:26

score 2 · Accepted Answer · answered Aug 11 '18 at 11:26

2

Since $\omega+\omega^2=-1$ and $\omega\times\omega^2=1$, your quadratic factor is $x^2-(\omega+\omega^2)yx+\omega\omega^2y^2=x^2+xy+y^2$, and mutiplication with $x-y$ gives $=x^3+x^2y+xy^2-x^2y-xy^2-y^3=x^3-y^3$ as desired.

answered Aug 11 '18 at 11:26

J.G.

115,835

Thanks, I see what I was missing. – Muhammad Salman Aug 11 '18 at 11:30

Solving roots of unity

1 Answers1