# Converting $ax^3+bx^2+cx+d$ to $t^3+pt+q$

Pub: Sep. 13, 2014 | Wri: a few years ago

$\def \l {\left} \def \r {\right} \def \f {\frac} ax^3+bx^2+cx+d = 0 \\ \def \x {t-\f{b}{3a}} \text{Let } x = \x \\ \def \bx {\l(\x\r)} a\bx^3 + b\bx^2 + c\bx + d = 0 \\ a\l(t^3-\f{b}{a}t^2+\f{b^2}{3a^2}t-\f{b^3}{27a^3}\r) + b\l(t^2-\f{2b}{3a}t+\f{b^2}{9a^2}\r) + c\l(t-\f{b}{3a}\r) + d = 0 \\ t^3-\f{b}{a}t^2+\f{b^2}{3a^2}t-\f{b^3}{27a^3} + \f{b}{a}t^2-\f{2b^2}{3a^2}t+\f{b^3}{9a^3} + \f{c}{a}t-\f{bc}{3a^2} + \f{d}{a} = 0 \\ t^3+\f{b^2}{3a^2}t-\f{2b^2}{3a^2}t+\f{c}{a}t - \f{b^3}{27a^3}+\f{b^3}{9a^3}-\f{bc}{3a^2}+\f{d}{a} = 0 \\ t^3+\l(\f{3ac-b^2}{3a^2}\r)t-\f{b^3}{27a^3}+\f{3b^3}{27a^3} - \f{9abc}{27a^3}+\f{27a^2d}{27a^3} = 0 \\ t^3+\l(\f{3ac-b^2}{3a^2}\r)t-\l(\f{2b^3-9abc+27a^2d}{27a^3}\r) = 0 \\ t^3+pt+q = 0 \\ \begin{split} p\, & = \f{3ac-b^2}{3a^2} \\ & = \f{c}{a}-\f{b}{a}\l(\f{b}{3a}\r) \end{split} \qquad \begin{split} q\, & = \f{2b^3-9abc+27a^2d}{27a^3} \\[3pt] & = 2\l(\f{b}{3a}\r)^3 - \f{c}{a}\l(\f{b}{3a}\r) + \f{d}{a} \end{split} \\~\\[3pt] \text{If a = 1, } \\[9pt] \begin{split} p\, & = \f{3c-b^2}{3} \\ & = c-b\l(\f{b}{3}\r) \end{split} \qquad \begin{split} q\, & = \f{2b^3-9bc+27d}{27} \\[3pt] & = 2\l(\f{b}{3}\r)^3 - c\l(\f{b}{3}\r) + d \end{split}$