Cubic Formula 1
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 \und#1{\style{text-decoration: underline}{#1}}
\def \link#1#2{\und{\href{#1}{#2}}}
\def \tlink#1#2{\link{#1}{\text{#2}}}
\def \lq {\unicode{x201C}}
\def \rq {\unicode{x201D}}
\def \qt#1{\lq #1 \rq}
\text{Note: This derivation is based on steps found at $\link{http://www.sosmath.com/algebra/factor/fac11/fac11.html}{\text{The }\qt{\text{Cubic Formula}}\text{ - SOS Math}}$ and other sites.} \\[6pt]
t^3+pt+q = 0 \\[6pt]
\text{Let $t = u+v$} \\[6pt]
(u+v)^3 + p(u+v) + q = 0 \\
u^3 + 3u^2v + 3uv^2 + v^3 + p(u+v) + q = 0 \\
u^3 + v^3 + 3uv(u+v) + p(u+v) + q = 0 \\
u^3 + v^3 + q + (3uv + p)(u+v) = 0 \\[6pt]
\begin{split}
& u^3 + v^3 + q = 0 \\
& u^3 + v^3 = -q
\end{split}
\qquad
\begin{split}
& 3uv + p = 0 \\
& 3uv = -p
\end{split} \\
v = -\f{p}{3u} \\
u^3 + \b{-\f{p}{3u}}^3 = -q \\
u^3 - \f{p^3}{27u^3} = -q \\
u^6 - \f{p^3}{27} = -qu^3 \\
u^6 + qu^3 - \f{p^3}{27} = 0 \\[12pt]
\begin{split}
u^3 &= \f{-q \pm \sqrt{q^2 + 4\b{\f{p^3}{27}}}}{2} \\[3pt]
&= \f{-q \pm 2\sqrt{\f{q^2}{4} + \f{p^3}{27}}}{2} \\[3pt]
&= -\f{q}{2} \pm \sqrt{\f{q^2}{4} + \f{p^3}{27}}
\end{split} \\[53pt]
\begin{split}
v^3 &= -q - u^3 \\[3pt]
&= -q + \f{q}{2} \mp \sqrt{\f{q^2}{4} + \f{p^3}{27}} \\[3pt]
&= -\f{q}{2} \mp \sqrt{\f{q^2}{4} + \f{p^3}{27}}
\end{split} \\[39pt]
\def \root [#1]#2{\sqrt[\leftroot{2}\uproot{2}\scriptstyle #1]{#2}}
\begin{split}
& u^3 = -\f{q}{2} + \sqrt{\f{q^2}{4} + \f{p^3}{27}} \\[3pt]
& \begin{split}
u &= \root[3]{-\f{q}{2} + \sqrt{\f{q^2}{4} + \f{p^3}{27}}} \\[3pt]
&= \root[3]{-\f{q}{2} + \sqrt{\b{\f{q}{2}}^2 + \b{\f{p}{3}}^3}}
\end{split}
\end{split}
\qquad
\begin{split}
& v^3 = -\f{q}{2} - \sqrt{\f{q^2}{4} + \f{p^3}{27}} \\[3pt]
& \begin{split}
v &= \root[3]{-\f{q}{2} - \sqrt{\f{q^2}{4} + \f{p^3}{27}}} \\[3pt]
&= \root[3]{-\f{q}{2} - \sqrt{\b{\f{q}{2}}^2 + \b{\f{p}{3}}^3}}
\end{split}
\end{split} \\[53pt]
\begin{split}
t_1 & = u+v \text{, take real cube root} \\
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} \\[33pt]
\text{For derivation of $t_2$ and $t_3$, see $\tlink{cubic-formula2.html}{Cubic Formula 2}$}
\]