Soit
a
x
3
+
b
x
2
+
c
x
+
d
=
0
{\displaystyle ax^{3}+bx^{2}+cx+d=0}
a
≠
0
{\displaystyle a\neq 0}
y
=
x
+
b
3
a
{\displaystyle y=x+{\frac {b}{3a}}}
a
{\displaystyle a}
y
3
+
p
y
+
q
=
0
{\displaystyle y^{3}+py+q=0}
{
p
=
3
a
c
−
b
2
3
a
2
q
=
2
b
3
−
9
a
b
c
+
27
a
2
d
27
a
3
{\displaystyle {\begin{cases}p={\frac {3ac-b^{2}}{3a^{2}}}\\q={\frac {2b^{3}-9abc+27a^{2}d}{27a^{3}}}\end{cases}}}
Transformation de Tschirnhaus
On a
a
(
y
−
b
3
a
)
3
+
b
(
y
−
b
3
a
)
2
+
c
(
y
−
b
3
a
)
+
d
=
0
{\displaystyle a\left(y-{\frac {b}{3a}}\right)^{3}+b\left(y-{\frac {b}{3a}}\right)^{2}+c\left(y-{\frac {b}{3a}}\right)+d=0}
soit
a
(
y
3
−
b
a
y
2
+
b
2
3
a
2
y
−
b
3
27
a
3
)
+
b
(
y
2
−
2
b
3
a
y
+
b
2
9
a
2
)
+
c
(
y
−
b
3
a
)
+
d
=
0
{\displaystyle a\left(y^{3}-{\frac {b}{a}}y^{2}+{\frac {b^{2}}{3a^{2}}}y-{\frac {b^{3}}{27a^{3}}}\right)+b\left(y^{2}-{\frac {2b}{3a}}y+{\frac {b^{2}}{9a^{2}}}\right)+c\left(y-{\frac {b}{3a}}\right)+d=0}
soit
a
y
3
+
(
a
b
2
−
2
a
b
2
+
3
a
2
c
3
a
2
)
y
+
(
−
a
b
3
+
3
a
b
3
−
9
a
2
b
c
+
27
a
3
d
27
a
3
)
=
0
{\displaystyle ay^{3}+\left({\frac {ab^{2}-2ab^{2}+3a^{2}c}{3a^{2}}}\right)y+\left({\frac {-ab^{3}+3ab^{3}-9a^{2}bc+27a^{3}d}{27a^{3}}}\right)=0}
soit
y
3
+
(
3
a
c
−
b
2
3
a
2
)
y
+
(
2
b
3
−
9
a
b
c
+
27
a
2
d
27
a
3
)
=
0
{\displaystyle y^{3}+\left({\frac {3ac-b^{2}}{3a^{2}}}\right)y+\left({\frac {2b^{3}-9abc+27a^{2}d}{27a^{3}}}\right)=0}
soit
y
3
+
p
y
+
q
=
0
{\displaystyle y^{3}+py+q=0}
.
En exprimant
y
=
y
α
+
y
β
{\displaystyle y=y_{\alpha }+y_{\beta }}
y
3
=
y
α
3
+
y
β
3
+
3
y
α
y
β
(
y
α
+
y
β
)
⏞
y
{\displaystyle y^{3}=y_{\alpha }^{3}+y_{\beta }^{3}+3y_{\alpha }y_{\beta }\overbrace {\left(y_{\alpha }+y_{\beta }\right)} ^{y}}
y
3
−
(
3
y
α
y
β
)
y
−
(
y
α
3
+
y
β
3
)
=
0
{\displaystyle y^{3}-\left(3y_{\alpha }y_{\beta }\right)y-\left(y_{\alpha }^{3}+y_{\beta }^{3}\right)=0}
Par identification, on a alors
{
−
3
y
α
y
β
=
p
−
y
α
3
−
y
β
3
=
q
{\displaystyle {\begin{cases}-3y_{\alpha }y_{\beta }=p\\-y_{\alpha }^{3}-y_{\beta }^{3}=q\end{cases}}}
Substitution de Viète )
y
β
=
−
p
3
y
α
{\displaystyle y_{\beta }=-{\frac {p}{3y_{\alpha }}}}
y
α
{\displaystyle y_{\alpha }}
y
β
{\displaystyle y_{\beta }}
On a alors
y
α
6
+
q
y
α
3
−
(
p
3
)
3
=
0
{\displaystyle y_{\alpha }^{6}+qy_{\alpha }^{3}-\left({\frac {p}{3}}\right)^{3}=0}
y
α
=
j
k
−
q
2
±
(
q
2
)
2
+
(
p
3
)
3
3
{\displaystyle y_{\alpha }=\mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}\pm {\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}
j
=
e
i
2
π
3
{\displaystyle \mathrm {j} =\mathrm {e} ^{\mathrm {i} {\frac {2\pi }{3}}}}
k
∈
{
0
;
1
;
2
}
{\displaystyle k\in \{0;1;2\}}
y
α
{\displaystyle y_{\alpha }}
j
k
−
q
2
+
(
q
2
)
2
+
(
p
3
)
3
3
×
j
3
−
k
−
q
2
−
(
q
2
)
2
+
(
p
3
)
3
3
=
−
p
3
{\displaystyle \mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}\times \mathrm {j} ^{3-k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}=-{\frac {p}{3}}}
y
k
=
j
k
−
q
2
+
(
q
2
)
2
+
(
p
3
)
3
3
+
j
3
−
k
−
q
2
−
(
q
2
)
2
+
(
p
3
)
3
3
{\displaystyle y_{k}=\mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}+\mathrm {j} ^{3-k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}
Cas où p et q sont réels
Si
p
{\displaystyle p}
q
{\displaystyle q}
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
d
{\displaystyle d}
Δ
=
−
(
4
p
3
+
27
q
2
)
{\displaystyle \Delta =-\left(4p^{3}+27q^{2}\right)}
Si
Δ
<
0
{\displaystyle \Delta <0}
y
3
{\displaystyle y_{3}}
y
1
{\displaystyle y_{1}}
y
2
{\displaystyle y_{2}}
y
k
=
−
j
k
q
2
+
(
q
2
)
2
+
(
p
3
)
3
3
−
j
3
−
k
q
2
−
(
q
2
)
2
+
(
p
3
)
3
3
{\displaystyle y_{k}=-\mathrm {j} ^{k}{\sqrt[{3}]{{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}-\mathrm {j} ^{3-k}{\sqrt[{3}]{{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}
Si
Δ
=
0
{\displaystyle \Delta =0}
Si
p
=
0
{\displaystyle p=0}
y
1
=
y
2
=
y
3
=
0
{\displaystyle y_{1}=y_{2}=y_{3}=0}
Si
p
≠
0
{\displaystyle p\neq 0}
y
1
=
y
2
=
−
3
q
2
p
{\displaystyle y_{1}=y_{2}=-{\frac {3q}{2p}}}
y
3
=
3
q
p
{\displaystyle y_{3}={\frac {3q}{p}}}
{
y
1
=
y
2
=
(
j
+
j
2
)
−
q
2
3
=
q
2
3
=
(
q
2
)
(
q
2
)
2
(
−
p
3
)
3
3
=
−
3
q
2
p
y
3
=
−
y
1
−
y
2
puisque
y
3
+
p
y
+
q
=
0
n'a pas de coefficient en
y
2
{\displaystyle {\begin{cases}y_{1}=y_{2}=\left(\mathrm {j} +\mathrm {j} ^{2}\right){\sqrt[{3}]{-{\frac {q}{2}}}}={\sqrt[{3}]{\frac {q}{2}}}={\sqrt[{3}]{\left({\frac {q}{2}}\right){\frac {\left({\frac {q}{2}}\right)^{2}}{\left({-{\frac {p}{3}}}\right)^{3}}}}}=-{\frac {3q}{2p}}\\y_{3}=-y_{1}-y_{2}{\text{ puisque }}y^{3}+py+q=0{\text{ n'a pas de coefficient en }}y^{2}\end{cases}}}
Si
Δ
>
0
{\displaystyle \Delta >0}
casus irreducibilis ), mais à partir de fonctions trigonométriques. On a
p
≠
0
{\displaystyle p\neq 0}
−
q
2
+
i
−
(
q
2
)
2
−
(
p
3
)
3
=
(
−
p
3
)
3
/
2
e
i
arccos
−
q
2
(
q
2
)
2
−
(
q
2
)
2
−
(
p
3
)
3
=
ρ
e
i
θ
{\displaystyle -{\frac {q}{2}}+\mathrm {i} {\sqrt {-\left({\frac {q}{2}}\right)^{2}-\left({\frac {p}{3}}\right)^{3}}}=\left({\frac {-p}{3}}\right)^{3/2}\mathrm {e} ^{\mathrm {i} \arccos {\frac {-{\frac {q}{2}}}{\sqrt {\left({\frac {q}{2}}\right)^{2}-\left({\frac {q}{2}}\right)^{2}-\left({\frac {p}{3}}\right)^{3}}}}}=\rho \mathrm {e} ^{\mathrm {i} \theta }}
{
ρ
=
(
−
p
3
)
3
2
θ
=
arccos
(
3
q
2
p
3
−
p
)
{\displaystyle {\begin{cases}\rho =\left({\frac {-p}{3}}\right)^{\frac {3}{2}}\\\theta =\arccos \left({{\frac {3q}{2p}}{\sqrt {\frac {3}{-p}}}}\right)\end{cases}}}
y
k
=
j
k
ρ
e
i
θ
3
+
j
3
−
k
ρ
e
−
i
θ
3
=
2
ρ
3
cos
θ
+
2
k
π
3
=
2
−
p
3
cos
[
arccos
(
3
q
2
p
3
−
p
)
3
+
2
k
π
3
]
{\displaystyle y_{k}=\mathrm {j} ^{k}{\sqrt[{3}]{\rho \mathrm {e} ^{\mathrm {i} \theta }}}+\mathrm {j} ^{3-k}{\sqrt[{3}]{\rho \mathrm {e} ^{-\mathrm {i} \theta }}}=2{\sqrt[{3}]{\rho }}\cos {\frac {\theta +2k\pi }{3}}=2{\sqrt {\frac {-p}{3}}}\cos {\Biggl [}{\frac {\arccos \left({{\frac {3q}{2p}}{\sqrt {\frac {3}{-p}}}}\right)}{3}}+{\frac {2k\pi }{3}}{\Biggr ]}}
Note : Pour calculer les 3 valeurs réelles, il nous a été nécessaire de calculer la racine carrée d'un nombre négatif. C'est Raphaël Bombelli qui a inventé
i
=
−
1
{\displaystyle \mathrm {i} ={\sqrt {-1}}}
On pose :
{
p
=
3
a
c
−
b
2
3
a
2
q
=
2
b
3
−
9
a
b
c
+
27
a
2
d
27
a
3
Δ
=
−
(
4
p
3
+
27
q
2
)
j
=
e
i
2
π
3
k
∈
{
0
;
1
;
2
}
{\displaystyle {\begin{cases}p={\frac {3ac-b^{2}}{3a^{2}}}\\q={\frac {2b^{3}-9abc+27a^{2}d}{27a^{3}}}\\\Delta =-\left(4p^{3}+27q^{2}\right)\\\mathrm {j} =\mathrm {e} ^{\mathrm {i} {\frac {2\pi }{3}}}\\k\in \{0;1;2\}\end{cases}}}
On a :
x
k
=
−
b
3
a
+
j
k
−
q
2
+
(
q
2
)
2
+
(
p
3
)
3
3
+
j
3
−
k
−
q
2
−
(
q
2
)
2
+
(
p
3
)
3
3
{\displaystyle x_{k}=-{\frac {b}{3a}}+\mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}+\mathrm {j} ^{3-k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}
c.-à-d. :
Si
Δ
<
0
{\displaystyle \Delta <0}
x
3
{\displaystyle x_{3}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
k
=
−
b
3
a
+
j
k
−
q
2
+
(
q
2
)
2
+
(
p
3
)
3
3
+
j
3
−
k
−
q
2
−
(
q
2
)
2
+
(
p
3
)
3
3
{\displaystyle x_{k}=-{\frac {b}{3a}}+\mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}+\mathrm {j} ^{3-k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}
Si
Δ
=
0
{\displaystyle \Delta =0}
Si
p
=
0
{\displaystyle p=0}
x
1
=
x
2
=
x
3
=
−
b
3
a
{\displaystyle x_{1}=x_{2}=x_{3}=-{\frac {b}{3a}}}
Si
p
≠
0
{\displaystyle p\neq 0}
x
1
=
x
2
=
−
b
3
a
−
3
q
2
p
{\displaystyle x_{1}=x_{2}=-{\frac {b}{3a}}-{\frac {3q}{2p}}}
x
3
=
−
b
3
a
+
3
q
p
{\displaystyle x_{3}=-{\frac {b}{3a}}+{\frac {3q}{p}}}
Si
Δ
>
0
{\displaystyle \Delta >0}
x
k
=
−
b
3
a
+
2
−
p
3
cos
[
arccos
(
3
q
2
p
3
−
p
)
3
+
2
k
π
3
]
{\displaystyle x_{k}=-{\frac {b}{3a}}+2{\sqrt {\frac {-p}{3}}}\cos {\Biggl [}{\frac {\arccos \left({{\frac {3q}{2p}}{\sqrt {\frac {3}{-p}}}}\right)}{3}}+{\frac {2k\pi }{3}}{\Biggr ]}}
a , b , c et d )
modifier
Finalement, les trois racines, qui peuvent dans certains cas être égales, sont :
{
x
1
=
−
b
3
a
−
1
3
a
2
b
3
−
9
a
b
c
+
27
a
2
d
+
(
2
b
3
−
9
a
b
c
+
27
a
2
d
)
2
−
4
(
b
2
−
3
a
c
)
3
2
3
−
1
3
a
2
b
3
−
9
a
b
c
+
27
a
2
d
−
(
2
b
3
−
9
a
b
c
+
27
a
2
d
)
2
−
4
(
b
2
−
3
a
c
)
3
2
3
x
2
=
−
b
3
a
+
1
+
i
3
6
a
2
b
3
−
9
a
b
c
+
27
a
2
d
+
(
2
b
3
−
9
a
b
c
+
27
a
2
d
)
2
−
4
(
b
2
−
3
a
c
)
3
2
3
+
1
−
i
3
6
a
2
b
3
−
9
a
b
c
+
27
a
2
d
−
(
2
b
3
−
9
a
b
c
+
27
a
2
d
)
2
−
4
(
b
2
−
3
a
c
)
3
2
3
x
3
=
−
b
3
a
+
1
−
i
3
6
a
2
b
3
−
9
a
b
c
+
27
a
2
d
+
(
2
b
3
−
9
a
b
c
+
27
a
2
d
)
2
−
4
(
b
2
−
3
a
c
)
3
2
3
+
1
+
i
3
6
a
2
b
3
−
9
a
b
c
+
27
a
2
d
−
(
2
b
3
−
9
a
b
c
+
27
a
2
d
)
2
−
4
(
b
2
−
3
a
c
)
3
2
3
{\displaystyle {\begin{cases}x_{1}=-{\frac {b}{3a}}-{\frac {1}{3a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}-{\frac {1}{3a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{2}=-{\frac {b}{3a}}+{\frac {1+\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}+{\frac {1-\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{3}=-{\frac {b}{3a}}+{\frac {1-\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}+{\frac {1+\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\end{cases}}}
(
2
b
3
−
9
a
b
c
+
27
a
2
d
)
2
−
4
(
b
2
−
3
a
c
)
3
=
0
⇒
{
3
a
c
−
b
2
=
0
⇒
x
1
=
x
2
=
x
3
=
−
b
3
a
3
a
c
−
b
2
≠
0
⇒
x
1
=
x
2
=
9
a
d
−
b
c
2
b
2
−
6
a
c
et
x
3
=
4
a
b
c
−
9
a
2
d
−
b
3
a
b
2
−
3
a
2
c
{\displaystyle \left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}=0\Rightarrow {\begin{cases}3ac-b^{2}=0\Rightarrow x_{1}=x_{2}=x_{3}=-{\frac {b}{3a}}\\3ac-b^{2}\neq 0\Rightarrow x_{1}=x_{2}={\frac {9ad-bc}{2b^{2}-6ac}}{\text{ et }}x_{3}={\frac {4abc-9a^{2}d-b^{3}}{ab^{2}-3a^{2}c}}\end{cases}}}
![{\displaystyle y_{\alpha }=\mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}\pm {\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09b635a8c1e4df2b757a837d9561cf9e4ee07cc4)
![{\displaystyle \mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}\times \mathrm {j} ^{3-k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}=-{\frac {p}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8958fae3cd211bdac68046e5c726871c8dbcc805)
![{\displaystyle y_{k}=\mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}+\mathrm {j} ^{3-k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd282075da6d3b583ce8d816d71fd839cc46079a)
![{\displaystyle y_{k}=-\mathrm {j} ^{k}{\sqrt[{3}]{{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}-\mathrm {j} ^{3-k}{\sqrt[{3}]{{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ea960271286e88dc2e2edd1426f67b1c40094fc)
![{\displaystyle {\begin{cases}y_{1}=y_{2}=\left(\mathrm {j} +\mathrm {j} ^{2}\right){\sqrt[{3}]{-{\frac {q}{2}}}}={\sqrt[{3}]{\frac {q}{2}}}={\sqrt[{3}]{\left({\frac {q}{2}}\right){\frac {\left({\frac {q}{2}}\right)^{2}}{\left({-{\frac {p}{3}}}\right)^{3}}}}}=-{\frac {3q}{2p}}\\y_{3}=-y_{1}-y_{2}{\text{ puisque }}y^{3}+py+q=0{\text{ n'a pas de coefficient en }}y^{2}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c038234701a81e344daa457bf811ca98d622fc6)
![{\displaystyle y_{k}=\mathrm {j} ^{k}{\sqrt[{3}]{\rho \mathrm {e} ^{\mathrm {i} \theta }}}+\mathrm {j} ^{3-k}{\sqrt[{3}]{\rho \mathrm {e} ^{-\mathrm {i} \theta }}}=2{\sqrt[{3}]{\rho }}\cos {\frac {\theta +2k\pi }{3}}=2{\sqrt {\frac {-p}{3}}}\cos {\Biggl [}{\frac {\arccos \left({{\frac {3q}{2p}}{\sqrt {\frac {3}{-p}}}}\right)}{3}}+{\frac {2k\pi }{3}}{\Biggr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8cfdf3fc9be4fb5df6e5b0ab6b55b139c1f4f42)
![{\displaystyle x_{k}=-{\frac {b}{3a}}+\mathrm {j} ^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}+\mathrm {j} ^{3-k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/224ebd8232a130ba87f9c7bf6ba4c33ea30250ce)
![{\displaystyle x_{k}=-{\frac {b}{3a}}+2{\sqrt {\frac {-p}{3}}}\cos {\Biggl [}{\frac {\arccos \left({{\frac {3q}{2p}}{\sqrt {\frac {3}{-p}}}}\right)}{3}}+{\frac {2k\pi }{3}}{\Biggr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f47785ebb8f0f7162e8e63993fb3c1ceddac580)
![{\displaystyle {\begin{cases}x_{1}=-{\frac {b}{3a}}-{\frac {1}{3a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}-{\frac {1}{3a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{2}=-{\frac {b}{3a}}+{\frac {1+\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}+{\frac {1-\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{3}=-{\frac {b}{3a}}+{\frac {1-\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}+{\frac {1+\mathrm {i} {\sqrt {3}}}{6a}}&{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d1ed356dd3ff9267e0a7592d874b936b48b7c76)
