Equação cúbica

Teorema de Scipione-Tartaglia

$a{x}^3+b{x}^2+cx+d=0$

Efetuando uma mudança de variável, da forma x = u + v, obtemos

$a(u+v{)}^3+b(u+v{)}^2+c(u+v)+d=0$

$a(u^3+3u^{2}v+3u{v}^2+v^3)+b(u^2+2uv+v^2)+c(u+v)+d=0$

$a{u}^3+3a{u}^{2}v+3au{v}^2+a{v}^3+b{u}^2+2buv+b{v}^2+cu+cv+d=0$

$a{u}^3+(3av+b)u^2+(3a{v}^2+2bv+c)u+a{v}^3+b{v}^2+cv+d=0$

$a{u}^3+(3av+b)u^2+(3a{v}^2+2bv+c)u+((av+b)v+c)v+d=0$

Neste passo tomamos o valor de v = $\frac{-b}{3a}$ para anular o termo com 'u²'

$a{u}^3+(3a\ast \frac{-b}{3a}+b)u^2+(3a(\frac{-b}{3a}{)}^2+2b\ast \frac{-b}{3a}+c)u+((a\ast \frac{-b}{3a}+b)\ast \frac{-b}{3a}+c)\ast \frac{-b}{3a}+d=0$

$a{u}^3+(\frac{b^2}{3a}-\frac{2b^2}{3a}+c)u+((\frac{-b}{3}+b)\ast \frac{-b}{3a}+c)\ast \frac{-b}{3a}+d=0$

$a{u}^3+(-\frac{b^2}{3a}+c)u+(\frac{2b}{3}\ast \frac{-b}{3a}+c)\ast \frac{-b}{3a}+d=0$

$a{u}^3+(-\frac{b^2}{3a}+c)u+\frac{2b}{3}\ast (\frac{-b}{3a}{)}^2-\frac{bc}{3a}+d=0$

$a{u}^3+(-\frac{b^2}{3a}+c)u+\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$

Agora substituímos o valor u = t + w

$a(t+w{)}^3+(-\frac{b^2}{3a}+c)\ast (t+w)+\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$

$a(t^3+3t^{2}w+3t{w}^2+w^3)+(-\frac{b^2}{3a}+c)\ast t+(-\frac{b^2}{3a}+c)\ast w+\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$

$a{t}^3+3a{t}^{2}w+3at{w}^2+a{w}^3+(-\frac{b^2}{3a}+c)\ast t+(-\frac{b^2}{3a}+c)\ast w+\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$

$a{t}^3+a{w}^3+(3atw+(-\frac{b^2}{3a}+c))(t+w)+\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$

$3atw=\frac{b^2}{3a}-ca{t}^3+a{w}^3=-\frac{2b^3}{27a^2}+\frac{bc}{3a}-d$

$tw=\frac{b^2}{9a^2}-\frac{c}{3a}{t}^3+w^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}$

$t=(\frac{b^2}{9a^2}-\frac{c}{3a})\ast \frac{1}{w}{t}^3+w^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}$

Substituindo t

$((\frac{b^2}{9a^2}-\frac{c}{3a})\ast \frac{1}{w}{)}^3+w^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}$

$(\frac{b^2}{9a^2}-\frac{c}{3a}{)}^3\ast \frac{1}{w^3}+w^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}$

$((\frac{b^2}{9a^2}-\frac{c}{3a}{)}^3\ast \frac{1}{w^3}+w^3)\ast w^3=(-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast w^3$

$w^6+(\frac{b^2}{9a^2}-\frac{c}{3a}{)}^3=(-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast w^3$

$w^6-(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast w^3+(-\frac{c}{3a}+\frac{b^2}{9a^2}{)}^3=0$

$w^6-(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast w^3+(-1\ast (\frac{c}{3a}-\frac{b^2}{9a^2}){)}^3=0$

$w^6-(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast w^3+(-1{)}^3\ast (\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3=0$

$w^6-(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast w^3-(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3=0$

$w^3=\frac{(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\pm \sqrt{(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}{)}^2+4(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}{2}$

$w^3=\frac{(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})}{2}\pm \sqrt{\frac{(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}{)}^2+4(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}{4}}$

$w^3=(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast \frac{1}{2}\pm \sqrt{(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}{)}^2\ast \frac{1}{4}+4(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3\ast \frac{1}{4}}$

$w^3=(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}{)}^2\ast (\frac{1}{2}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$w^3=(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}\ast (\frac{1}{2}){)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$w^3=(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$w=\sqrt[3]{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}$

Substituímos o valor de w³ em:

$t^3+w^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}$

$t^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}-w^3$

$t^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}-(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$t^3=(-\frac{2b^3}{27a^3}+\frac{b^3}{27a^3}+\frac{bc}{3a^2}-\frac{bc}{6a^2}-\frac{d}{a}+\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$t^3=(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$t=\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}$

Sabendo que u = t + w e x = u + v, logo x = t + w + v, substituindo obtemos a fórmula de resolução da equação cúbica:

$x=-\frac{b}{3a}+\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}+\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}$

Ou ainda, fazendo algumas simplificações:

$x=-\frac{b}{3a}+\sqrt[3]{((-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast \frac{1}{2})\pm \sqrt{((-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast \frac{1}{2}{)}^2+((\frac{c}{a}-\frac{b^2}{3a^2})\ast \frac{1}{3}{)}^3}}+\sqrt[3]{((-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast \frac{1}{2})\pm \sqrt{((-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast \frac{1}{2}{)}^2+((\frac{c}{a}-\frac{b^2}{3a^2})\ast \frac{1}{3}{)}^3}}$

$x=-\frac{b}{3a}+\sqrt[3]{(-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast \frac{1}{2}\pm \sqrt{(-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}{)}^2\ast \frac{1}{4}+(\frac{c}{a}-\frac{b^2}{3a^2}{)}^3\ast \frac{1}{27}}}+\sqrt[3]{(-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast \frac{1}{2}\pm \sqrt{(-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}{)}^2\ast \frac{1}{4}+(\frac{c}{a}-\frac{b^2}{3a^2}{)}^3\ast \frac{1}{27}}}$

como há várias partes que se repetem na fórmula, para simplificar podemos fazer:

$r=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}s=\frac{c}{a}-\frac{b^2}{3a^2}$

$x=-\frac{b}{3a}+\sqrt[3]{\frac{r}{2}\pm \sqrt{\frac{r^2}{4}+\frac{s^3}{27}}}+\sqrt[3]{\frac{r}{2}\pm \sqrt{\frac{r^2}{4}+\frac{s^3}{27}}}$

ou ainda para obter a fórmula clássica fazemos:

$q=\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}p=\frac{c}{a}-\frac{b^2}{3a^2}$

$x=-\frac{b}{3a}+\sqrt[3]{-\frac{q}{2}\pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}\pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$

para obter a fórmula clássica diretamente, fazemos:

$w^6-(\frac{-2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a})\ast w^3-(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3=0$

$w^6+(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast w^3-(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3=0$

$w^3=\frac{-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\pm \sqrt{(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}{)}^2+4(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}{2}$

$w^3=\frac{-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})}{2}\pm \sqrt{\frac{(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}{)}^2+4(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}{4}}$

$w^3=-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast \frac{1}{2}\pm \sqrt{(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}{)}^2\ast \frac{1}{4}+4(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3\ast \frac{1}{4}}$

$w^3=-(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}{)}^2\ast (\frac{1}{2}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$w^3=-(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}\ast (\frac{1}{2}){)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$w^3=-(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$w=\sqrt[3]{-(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}$

Substituímos o valor de w³ em:

$t^3+w^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}$

$t^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}-w^3$

$t^3=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}+(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$t^3=(-\frac{2b^3}{27a^3}+\frac{b^3}{27a^3}+\frac{bc}{3a^2}-\frac{bc}{6a^2}-\frac{d}{a}+\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$t^3=(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}$

$t=\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}$

$t=\sqrt[3]{-(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}$

Sabendo que u = t + w e x = u + v, logo x = t + w + v, substituindo obtemos a fórmula de resolução da equação cúbica:

$x=-\frac{b}{3a}+\sqrt[3]{-(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}+\sqrt[3]{-(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a})\pm \sqrt{(\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}{)}^2+(\frac{c}{3a}-\frac{b^2}{9a^2}{)}^3}}$

Ou ainda, fazendo algumas simplificações:

$x=-\frac{b}{3a}+\sqrt[3]{(-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast \frac{1}{2})\pm \sqrt{(-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast \frac{1}{2}{)}^2+((\frac{c}{a}-\frac{b^2}{3a^2})\ast \frac{1}{3}{)}^3}}+\sqrt[3]{(-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast \frac{1}{2})\pm \sqrt{(-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast \frac{1}{2}{)}^2+((\frac{c}{a}-\frac{b^2}{3a^2})\ast \frac{1}{3}{)}^3}}$

$x=-\frac{b}{3a}+\sqrt[3]{-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast \frac{1}{2}\pm \sqrt{-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}{)}^2\ast \frac{1}{4}+(\frac{c}{a}-\frac{b^2}{3a^2}{)}^3\ast \frac{1}{27}}}+\sqrt[3]{-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})\ast \frac{1}{2}\pm \sqrt{-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}{)}^2\ast \frac{1}{4}+(\frac{c}{a}-\frac{b^2}{3a^2}{)}^3\ast \frac{1}{27}}}$

como há várias partes que se repetem na fórmula, para simplificar podemos fazer:

$r=-\frac{2b^3}{27a^3}+\frac{bc}{3a^2}-\frac{d}{a}s=\frac{c}{a}-\frac{b^2}{3a^2}$

$x=-\frac{b}{3a}+\sqrt[3]{\frac{r}{2}\pm \sqrt{\frac{r^2}{4}+\frac{s^3}{27}}}+\sqrt[3]{\frac{r}{2}\pm \sqrt{\frac{r^2}{4}+\frac{s^3}{27}}}$

ou ainda para obter a fórmula clássica fazemos:

$q=\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}p=\frac{c}{a}-\frac{b^2}{3a^2}$

$x=-\frac{b}{3a}+\sqrt[3]{-\frac{q}{2}\pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}\pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$