XoaX.net's MathML: The Solution of the Cubic Equation

The solution of the cubic equation can be found by reducing the equation as follows:

\begin{matrix} 0 & = & a x^{3} + b x^{2} + c x + d \\ = & a y^{3} + (c - \frac{b^{2}}{3 a}) y + (d + \frac{2 b^{3}}{27 a^{2}} - \frac{b c}{3 a}) \\ = & a (y^{3} + (\frac{c}{a} - \frac{b^{2}}{3 a^{2}}) y + (\frac{d}{a} + \frac{2 b^{3}}{27 a^{3}} - \frac{b c}{3 a^{2}})) \\ = & a (y^{3} + (3 s t) y + (s^{3} - t^{3})) \end{matrix}

where

\begin{matrix} x & = & y - \frac{b}{3 a} \\ 3 s t & = & \frac{c}{a} - \frac{b^{2}}{3 a^{2}} \\ s^{3} - t^{3} & = & \frac{d}{a} + \frac{2 b^{3}}{27 a^{3}} - \frac{b c}{3 a^{2}} \end{matrix}

and $y = t - s$ will give us a solution to the equation. We can take the third equation and solve for $t$ , to get

t = \frac{1}{3 s} (\frac{c}{a} - \frac{b^{2}}{3 a^{2}})

Cubing both sides of the equation, we get

t^{3} = \frac{1}{s^{3}} {(\frac{c}{3 a} - \frac{b^{2}}{9 a^{2}})}^{3}

We can substitute this into the fourth equation and multiple by $s^{3}$ to get a quadratic in $s^{3}$

{(s^{3})}^{2} - (\frac{d}{a} + \frac{2 b^{3}}{27 a^{3}} - \frac{b c}{3 a^{2}}) s^{3} - {(\frac{c}{3 a} - \frac{b^{2}}{9 a^{2}})}^{3} = 0

Multiplying by ${(3 a)}^{6}$ , we get a quadratic in ${(3 a s)}^{3}$

{({(3 a s)}^{3})}^{2} - (3^{6} a^{5} d + 3^{3} a^{3} 2 b^{3} - 3^{5} a^{4} b c) {(3 a s)}^{3} - {(3 a c - b^{2})}^{3} = 0

This can be solved with the quadratic formula to get

{(3 a s)}^{3} = \frac{(3^{6} a^{5} d + 3^{3} a^{3} 2 b^{3} - 3^{5} a^{4} b c) \pm \sqrt{{(3^{6} a^{5} d + 3^{3} a^{3} 2 b^{3} - 3^{5} a^{4} b c)}^{2} + 4 {(3 a c - b^{2})}^{3}}}{2}

This we can easily solve for $s$ by taking a cubed root and dividing by $3 a$

s = \frac{1}{3 a} \sqrt[3]{\frac{(3^{6} a^{5} d + 3^{3} a^{3} 2 b^{3} - 3^{5} a^{4} b c) \pm \sqrt{{(3^{6} a^{5} d + 3^{3} a^{3} 2 b^{3} - 3^{5} a^{4} b c)}^{2} + 4 {(3 a c - b^{2})}^{3}}}{2}}

From the value of $s$ , we can get the value of $t$ . We make the observation that since $y = t - s$ is a root of our equation above, we can factor as follows

\begin{matrix} y^{3} + (3 s t) y + (s^{3} - t^{3}) & = & (y + s - t) (y^{2} + (t - s) y + s^{2} + s t + t^{2}) \end{matrix}

Now, the second quadratic can be factored via the quadratic formula to get

y = \frac{s - t \pm \sqrt{- 3 {(s + t)}^{2}}}{2} = \frac{s - t \pm i \sqrt{3} (s + t)}{2}