# Galois-theoretic derivation of the cubic formula

We are trying to find the roots ${r}_{1},{r}_{2},{r}_{3}$ of the polynomial^{} ${x}^{3}+a{x}^{2}+bx+c=0$. From the equation

$$(x-{r}_{1})(x-{r}_{2})(x-{r}_{3})={x}^{3}+a{x}^{2}+bx+c$$ |

we see that

$a$ | $=$ | $-({r}_{1}+{r}_{2}+{r}_{3})$ | ||

$b$ | $=$ | ${r}_{1}{r}_{2}+{r}_{1}{r}_{3}+{r}_{2}{r}_{3}$ | ||

$c$ | $=$ | $-{r}_{1}{r}_{2}{r}_{3}$ |

The goal is to explicitly construct a radical tower over the field $k=\u2102(a,b,c)$ that contains the three roots ${r}_{1},{r}_{2},{r}_{3}$.

Let $L=\u2102({r}_{1},{r}_{2},{r}_{3})$. By Galois theory^{} we know that
$\mathrm{Gal}(L/\u2102(a,b,c))={S}_{3}$. Let $K\subset L$ be the fixed field of
${A}_{3}\subset {S}_{3}$. We have a tower of field extensions