Equação cúbica

Teorema de Scipione-Tartaglia

$ax^{3}+bx^{2}+cx+d=0$

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

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

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

$au^{3}+3au^{2}v+3auv^{2}+av^{3}+bu^{2}+2buv+bv^{2}+cu+cv+d=0$

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

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

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

$au^{3}+(3a*{-b \over 3a}+b)u^{2}+(3a({-b \over 3a})^{2}+2b*{-b \over 3a}+c)u+((a*{-b \over 3a}+b)*{-b \over 3a}+c)*{-b \over 3a}+d=0$

$au^{3}+({b^{2} \over 3a}-{2b^{2} \over 3a}+c)u+(({-b \over 3}+b)*{-b \over 3a}+c)*{-b \over 3a}+d=0$

$au^{3}+(-{b^{2} \over 3a}+c)u+({2b \over 3}*{-b \over 3a}+c)*{-b \over 3a}+d=0$

$au^{3}+(-{b^{2} \over 3a}+c)u+{2b \over 3}*({-b \over 3a})^{2}-{bc \over 3a}+d=0$

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

Agora substituímos o valor u = t + w

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

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

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

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

$3atw={b^{2} \over 3a}-c\qquad \qquad at^{3}+aw^{3}=-{2b^{3} \over 27a^{2}}+{bc \over 3a}-d$

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

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

Substituindo t

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Substituímos o valor de w³ em:

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

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

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

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

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

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

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

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

Ou ainda, fazendo algumas simplificações:

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

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

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

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

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

ou ainda para obter a fórmula clássica fazemos:

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

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

para obter a fórmula clássica diretamente, fazemos:

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

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

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

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

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

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

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

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

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

Substituímos o valor de w³ em:

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

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

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

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

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

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

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

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

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

Ou ainda, fazendo algumas simplificações:

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

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

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

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

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

ou ainda para obter a fórmula clássica fazemos:

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

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