We are trying to find the roots $r_1, r_2, r_3$ of the polynomial $x^3 + ax^2 + bx + c = 0$ . From the equation $$ (x-r_1)(x-r_2)(x-r_3) = x^3 + ax^2 + bx + c $$ we see that \begin{eqnarray*} 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 \end{eqnarray*}The goal is to explicitly construct a radical tower over the field $k = \C(a,b,c)$ that contains the three roots $r_1, r_2, r_3$ .

Let $L = \C(r_1,r_2,r_3)$ . By Galois theory we know that $\Gal(L/\C(a,b,c)) = S_3$ . Let $K \subset L$ be the fixed field of $A_3 \subset S_3$ . We have a tower of field extensions

Let $\sigma := (123)$ be a generator of $\Gal(L/K) = A_3$ . Let $\omega = e^{2 \pi i/3} \in \C \subset L$ be a primitive cube root of unity. Since $\omega$ has norm 1, Hilbert's Theorem 90 tells us that $\omega = y/\sigma(y)$ for some $y \in L$ . Galois theory (or Kummer theory) then tells us that $L = K(y)$ and $y^3 \in K$ , thus exhibiting $L$ as a radical extension of $K$ .

The proof of Hilbert's Theorem 90 provides a procedure for finding $y$ , which is as follows: choose any $x \in L$ , form the quantity $$ \omega x + \omega^2 \sigma(x) + \omega^3 \sigma^2(x); $$ then this quantity automatically yields a suitable value for $y$ provided that it is nonzero. In particular, choosing $x = r_2$ yields $$ y = r_1 + \omega r_2 + \omega^2 r_3. $$ and we have $L = K(y)$ with $y^3 \in K$ . Moreover, since $\tau := (23)$ does not fix $y^3$ , it follows that $y^3 \notin k$ , and this, combined with $[K:k] = 2$ , shows that $K = k(y^3)$ .

Set $z := \tau(y) = r_1 + \omega^2 r_2 + \omega r_3$ . Applying the same technique to the extension $K/k$ , we find that $K = k(y^3 - z^3)$ with $(y^3-z^3)^2 \in k$ , and this exhibits $K$ as a radical extension of $k$ .

To get explicit formulas, start with $y^3 + z^3$ and $y^3 z^3$ , which are fixed by $S_3$ and thus guaranteed to be in $k$ . Using the reduction algorithm for symmetric polynomials, we find \begin{eqnarray*} y^3 + z^3 & = & -2a^3 + 9ab - 27c \\ y^3 z^3 & = & (a^2 - 3b)^3 \end{eqnarray*}Solving this system for $y$ and $z$ yields \begin{eqnarray*} y & = & \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{2}\right)^{1/3} \\ z & = & \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{2}\right)^{1/3} \end{eqnarray*}Now we solve the linear system \begin{eqnarray*} a & = & -(r_1+r_2+r_3) \\ y & = & r_1 + \omega r_2 + \omega^2 r_3 \\ z & = & r_1 + \omega^2 r_2 + \omega r_3 \end{eqnarray*}and we get \begin{eqnarray*} r_1 & = & \frac{1}{3} (-a + y + z) \\ r_2 & = & \frac{1}{3} (-a + \omega^2 y + \omega z) \\ r_3 & = & \frac{1}{3} (-a + \omega y + \omega^2 z) \end{eqnarray*}which expresses $r_1, r_2, r_3$ as radical expressions of $a,b,c$ by way of the previously obtained expressions for $y$ and $z$ , and completes the derivation of the cubic formula.