Cubic Formula 2
Pub: Sep. 13, 2014 | Wri: a few years ago
\[
\def \l {\left}
\def \r {\right}
\def \f {\frac}
\def \b#1{\l(#1\r)}
\def \root [#1]#2{\sqrt[\leftroot{2}\uproot{2}\scriptstyle #1]{#2}}
t^3+pt+q = 0 \\
t_1 = u+v
\qquad
p = -3uv
\qquad
q = -(u^3+v^3) \\
t^3 - 3uvt - (u^3+v^3) = 0 \\[6pt]
\def \p {\phantom}
\require{enclose}
\begin{split}
& \kern1.1ex t^2 + (u+v)t + (u^2-uv+v^2) \\[-2pt]
t-(u+v) & \enclose{longdiv}{t^3-3uvt-(u^3+v^3)\hspace{8.8ex}} \\[-2pt]
& \kern0.7ex
\begin{split}
& t^3-(u+v)t^2 \\[-2pt]
& \overline{\p{t^3-\;}(u+v)t^2}-3uvt \\[-2pt]
& \p{t^3-\;}(u+v)t^2-(u+v)^2t \\[-2pt]
& \p{t^3-\;}\overline{\p{(u+v)t^2-\;}(u^2+2uv+v^2)t}-3uvt \\[-2pt]
& \p{t^3-\;}\p{(u+v)t^2-\;}(u^2-uv+v^2)t-(u^3+v^3) \\[-2pt]
& \p{t^3-\;}\p{(u+v)t^2-\;}
\underline{(u^2-uv+v^2)t-(u+v)(u^2-uv+v^2)} \\[-2pt]
& \p{t^3-\;}\p{(u+v)t^2-\;}0 \\[-2pt]
\end{split}
\end{split} \\
t^2 + (u+v)t + u^2-uv+v^2 = 0 \\
\begin{split}
t &= \f{-(u+v) \pm \sqrt{u^2+2uv+v^2-4u^2+4uv-4v^2}}{2} \\
&= \f{-(u+v) \pm \sqrt{-3u^2+6uv-3v^2}}{2} \\
&= \f{-(u+v) \pm \sqrt{-3(u-v)^2}}{2} \\
&= -\f{1}{2}(u+v) \pm \f{\sqrt{3}}{2}\sqrt{-(u-v)^2}
\end{split} \\~\\[3pt]
\text{$u-v$ can only be real or purely imaginary.} \\[9pt]
\text{If $u-v$ is real,} \\
\sqrt{-(u-v)^2} = (u-v)i \\
\text{If $u-v$ is imaginary,} \\
\sqrt{-(u-v)^2} = -(u-v)i \\[15pt]
\text{This amounts to $t_2$ and $t_3$ switching places if we let} \\
\sqrt{-(u-v)^2} = (u-v)i \\[15pt]
\begin{split}
t_1 & = u+v \\
t_2 & = -\f{1}{2}(u+v) + \f{\sqrt{3}}{2}(u-v)i \\
t_3 & = -\f{1}{2}(u+v) - \f{\sqrt{3}}{2}(u-v)i
\end{split} \\
\]