cubic formula

The three roots $r_1, r_2, r_3$ of a cubic polynomial equation $x^3 + ax^2 + bx + c = 0$ are given by \begin{eqnarray*} r_1 & = & -\frac{a}{3} + \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ & & {} + \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ r_2 & = & -\frac{a}{3} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ & & {} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ r_3 & = & -\frac{a}{3} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\ & & {} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \end{eqnarray*}