Notes for Elements of Abstract Algebra

Introduction

$x^3+qx-r=0$

Let $x=u+v \implies (u^3+3u^2v+3uv^2+v^3)+q(u+v)-r=0$ $\implies u^3+v^3+(3uv+q)(u+v)-r=0$

Choose $u,v$ so that $3uv+q=0 \implies uv=-\dfrac{q}{3} \implies v=-\dfrac{q}{3u}$

$\implies u^6-ru^3-\dfrac{q^3}{27}=0 \implies u^3=\dfrac{r}{2} \pm \sqrt{\dfrac{r^2}{4}+\dfrac{q^3}{27}}$

Let $\omega=e^{i\tfrac{\pi}{3}}=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}, \omega^2=e^{-i\tfrac{\pi}{3}} = -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$, then three roots of original equation are:

$u+v, \omega u + \omega ^2v, \omega ^2 u + \omega v$

Chapter 1 - Set Theory