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}$} \]