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Posmatrajmo polinom trećeg stepena
y
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
d
{\displaystyle y(x)=ax^{3}+bx^{2}+cx+d}
a
≠
0
{\displaystyle a\neq 0}
koja siječe x-osu u tačkama koje su nule odgovarajuće jednačine trećeg stepena (kubne jednačine).
a
x
3
+
b
x
2
+
c
x
+
d
=
0
{\displaystyle ax^{3}+bx^{2}+cx+d=0}
Rješenja
x
1
,
x
2
,
x
3
{\displaystyle x_{1},x_{2},x_{3}}
a
x
3
+
b
x
2
+
c
x
+
d
=
0
{\displaystyle ax^{3}+bx^{2}+cx+d=0}
Vieteovih formula
x
1
+
x
2
+
x
3
=
−
b
a
,
x
1
x
2
+
x
2
x
3
+
x
3
x
1
=
c
a
,
x
1
x
2
x
3
=
−
d
a
.
{\displaystyle x_{1}+x_{2}+x_{3}=-{\frac {b}{a}},\ x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}={\frac {c}{a}},\ x_{1}x_{2}x_{3}=-{\frac {d}{a}}.}
[1] Primjer
x
3
−
3
x
2
−
x
+
3
=
0
{\displaystyle x^{3}-3x^{2}-x+3=0}
i ima nule, rješenja
x
=
−
1
{\displaystyle x=-1}
x
=
1
{\displaystyle x=1}
x
=
3
{\displaystyle x=3}
U jednačini
a
x
3
+
b
x
2
+
c
x
+
d
=
0
{\displaystyle ax^{3}+bx^{2}+cx+d=0}
a
≠
0
{\displaystyle a\neq 0}
smjenom
y
=
b
−
b
3
a
{\displaystyle y=b-{\frac {b}{3a}}}
postaje
y
3
+
p
y
+
q
=
0
{\displaystyle y^{3}+py+q=0}
za
p
=
c
a
−
b
2
3
a
2
{\displaystyle p={\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}}}
q
=
d
a
−
b
c
3
a
2
+
2
b
3
27
a
3
{\displaystyle q={\frac {d}{a}}-{\frac {bc}{3a^{2}}}+{\frac {2b^{3}}{27a^{3}}}}
Diskriminanta kubne jednačine [ uredi | uredi izvor ] Često se diskriminantom kubne jednačine naziva diskriminanta [2]
D
=
a
4
(
x
1
−
x
2
)
2
(
x
2
−
x
3
)
2
(
x
3
−
x
1
)
2
)
{\displaystyle D=a^{4}(x_{1}-x_{2})^{2}(x_{2}-x_{3})^{2}(x_{3}-x_{1})^{2})}
pripadnog polinoma :
f
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
d
{\displaystyle f(x)=ax^{3}+bx^{2}+cx+d}
x
1
,
x
2
,
x
3
{\displaystyle x_{1},x_{2},x_{3}}
D
=
−
4
b
3
d
+
b
2
c
2
−
4
a
c
3
+
18
a
b
c
d
−
27
a
2
d
2
.
{\displaystyle D=-4b^{3}d+b^{2}c^{2}-4ac^{3}+18abcd-27a^{2}d^{2}.}
Ovo vrijedi za sve kubne jednačine, a ne samo za one s realnim koeficijentima.
I izraz
D
=
q
2
4
+
p
3
27
{\displaystyle D={\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}
Za
D
<
0
{\displaystyle D<0}
Za
D
=
0
{\displaystyle D=0}
Za
D
>
0
{\displaystyle D>0}
Primjer Jednačina
y
=
x
3
−
3
x
2
−
x
+
3
{\displaystyle y=x^{3}-3x^{2}-x+3}
y
=
x
+
1
{\displaystyle y=x+1}
postaje
y
3
−
4
y
=
0
{\displaystyle y^{3}-4y=0}
Jednačina
y
3
+
p
y
+
q
=
0
{\displaystyle y^{3}+py+q=0}
y
=
u
+
v
{\displaystyle y=u+v}
postaje
u
3
+
v
3
+
(
3
u
v
+
p
)
(
u
+
v
)
+
q
=
0
{\displaystyle u^{3}+v^{3}+\left(3uv+p\right)\left(u+v\right)+q=0}
Ovdje smo nepoznatu
y
{\displaystyle y}
u
{\displaystyle u}
v
{\displaystyle v}
uvodimo novi uslov
3
u
v
=
p
{\displaystyle 3uv=p}
pa je
u
3
+
v
3
=
−
q
;
u
v
=
−
p
3
{\displaystyle u^{3}+v^{3}=-q;uv={\frac {-p}{3}}}
Iz
y
3
−
4
y
=
0
{\displaystyle y^{3}-4y=0}
dobijamo
u
3
+
v
3
=
0
;
u
v
=
4
3
{\displaystyle u^{3}+v^{3}=0;uv={\frac {4}{3}}}
Sistem
u
3
+
v
3
=
−
q
;
u
v
=
−
p
3
{\displaystyle u^{3}+v^{3}=-q;uv={\frac {-p}{3}}}
u
3
+
v
3
=
−
q
;
u
3
v
3
=
−
p
3
27
{\displaystyle u^{3}+v^{3}=-q;u^{3}v^{3}={\frac {-p^{3}}{27}}}
Iz kojeg dobijamo
t
2
+
q
t
−
−
p
3
27
{\displaystyle t^{2}+qt-{\frac {-p^{3}}{27}}}
za
t
1
=
u
3
i
t
2
=
v
3
{\displaystyle t_{1}=u^{3}it_{2}=v^{3}}
Njenim rješavanjem dobijamo
t
1
,
2
=
−
q
27
±
(
q
2
4
)
+
p
3
27
{\displaystyle t_{1},2=-{\frac {q}{27}}\pm {\sqrt {({\frac {q^{2}}{4}})+{\frac {p^{3}}{27}}}}}
Vračanjem druge smjene dobijamo
y
=
−
q
2
+
q
2
4
+
p
3
27
3
+
−
q
2
−
q
2
4
+
p
3
27
3
{\displaystyle y={\sqrt[{3}]{{\frac {-q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{{\frac {-q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}}
IZ
y
3
−
4
y
=
0
{\displaystyle y^{3}-4y=0}
p
=
−
4
i
q
=
0
{\displaystyle p=-4iq=0}
pa je
y
=
0
{\displaystyle y=0}
tj. jedno rješenje jednačine je
y
=
x
3
−
3
x
2
−
x
+
3
{\displaystyle y=x^{3}-3x^{2}-x+3}
x
=
o
{\displaystyle x=o}
Parovi (u,v) su rješenja sistema
u
3
+
v
3
=
−
q
;
u
v
=
−
p
3
{\displaystyle u^{3}+v^{3}=-q;uv={\frac {-p}{3}}}
Iz jednačina
u
3
=
t
1
i
v
3
=
t
2
{\displaystyle u^{3}=t_{1}iv^{3}=t_{2}}
slijedi
u
1
=
t
3
1
{\displaystyle u_{1}={\sqrt[{3}]{t}}_{1}}
u
2
=
a
t
3
1
{\displaystyle u_{2}=a{\sqrt[{3}]{t}}_{1}}
u
3
=
a
2
t
3
1
{\displaystyle u_{3}=a^{2}{\sqrt[{3}]{t}}_{1}}
v
1
=
t
3
2
{\displaystyle v_{1}={\sqrt[{3}]{t}}_{2}}
v
2
=
a
t
3
2
{\displaystyle v_{2}=a{\sqrt[{3}]{t}}_{2}}
v
3
=
a
2
t
3
2
{\displaystyle v_{3}=a^{2}{\sqrt[{3}]{t}}_{2}}
gdje su treči korjeni jedinice
a
0
=
1
,
a
1
=
1
2
(
−
1
+
i
3
3
)
,
a
2
=
1
2
(
−
1
−
i
3
3
)
{\displaystyle a^{0}=1,a^{1}={\frac {1}{2}}(-1+i{\sqrt[{3}]{3}}),a^{2}={\frac {1}{2}}(-1-i{\sqrt[{3}]{3}})}
Odnosno sva rješenja jednačine
y
3
+
p
y
+
q
=
0
{\displaystyle y^{3}+py+q=0}
data su formulom
y
1
=
t
3
1
+
t
3
2
{\displaystyle y_{1}={\sqrt[{3}]{t}}_{1}+{\sqrt[{3}]{t}}_{2}}
y
2
=
a
t
3
1
+
a
t
3
2
{\displaystyle y_{2}=a{\sqrt[{3}]{t}}_{1}+a{\sqrt[{3}]{t}}_{2}}
y
3
=
a
2
t
3
1
+
t
3
2
{\displaystyle y_{3}=a^{2}{\sqrt[{3}]{t}}_{1}+{\sqrt[{3}]{t}}_{2}}
koje se također nazivaju Kardanove formule.
U našem slučaju (za
D
<
0
{\displaystyle D<0}
D
<
0
{\displaystyle D<0}
Za
p
=
{\displaystyle p=}
q
=
0
{\displaystyle q=0}
D
=
−
64
27
{\displaystyle D={\frac {-64}{27}}}
pa je
y
=
0
{\displaystyle y=0}
y
2
=
a
t
3
1
+
a
t
3
2
{\displaystyle y_{2}=a{\sqrt[{3}]{t}}_{1}+a{\sqrt[{3}]{t}}_{2}}
y
3
=
a
2
t
3
1
+
t
3
2
{\displaystyle y_{3}=a^{2}{\sqrt[{3}]{t}}_{1}+{\sqrt[{3}]{t}}_{2}}
Iz ovoga je
t
1
=
−
t
2
{\displaystyle t_{1}=-t_{2}}
y
2
=
(
a
−
a
2
)
−
64
27
3
{\displaystyle y_{2}=(a-a^{2}){\sqrt[{3}]{\sqrt {\frac {-64}{27}}}}}
y
3
=
(
a
2
−
a
)
−
64
27
3
{\displaystyle y_{3}=(a^{2}-a){\sqrt[{3}]{\sqrt {\frac {-64}{27}}}}}
y
2
=
(
i
3
3
)
2
i
3
3
{\displaystyle y_{2}=(i{\sqrt[{3}]{3}}){\frac {2i}{\sqrt[{3}]{3}}}}
y
3
=
(
−
i
3
3
)
2
i
3
3
{\displaystyle y_{3}=(-i{\sqrt[{3}]{3}}){\frac {2i}{\sqrt[{3}]{3}}}}
y
2
=
2
∧
y
3
=
−
2
{\displaystyle y_{2}=2\land y_{3}=-2}
x
=
y
+
1
{\displaystyle x=y+1}
x
1
=
−
1
{\displaystyle x_{1}=-1}
x
2
=
1
,
{\displaystyle x_{2}=1,}
x
3
=
3
{\displaystyle x_{3}=3}
Opće rješenje za svaku kubnu jednačinu
a
x
3
+
b
x
2
+
c
x
+
d
=
0
{\displaystyle ax^{3}+bx^{2}+cx+d=\,0}
je
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{aligned}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+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-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-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+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{aligned}}}
Rešivost algebarskih jednačina Beograd 2011.
^ Vietove formule ^ Diskriminanta kubne jednačine 
![{\displaystyle y={\sqrt[{3}]{{\frac {-q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{{\frac {-q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3ec61fe60abed23b474dc87be6605fb30e9dbb9)
![{\displaystyle u_{1}={\sqrt[{3}]{t}}_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1aaa56fdcf5b05b90d6e1d318cc67e5ff8a7eaf4)
![{\displaystyle u_{2}=a{\sqrt[{3}]{t}}_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d30a6a142c4c644d9db3f5268fe16e294fbaccd)
![{\displaystyle u_{3}=a^{2}{\sqrt[{3}]{t}}_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af023ac3fdacc843cb206b2fc1edad780d2346a9)
![{\displaystyle v_{1}={\sqrt[{3}]{t}}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/663dcb723f2313a445410ec68a0879034da59b09)
![{\displaystyle v_{2}=a{\sqrt[{3}]{t}}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3b83ff5a661ecbff1fc1012f21e0674563ef7f3)
![{\displaystyle v_{3}=a^{2}{\sqrt[{3}]{t}}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1b23ae32e1dc56c905fc0685499f3ded1c5e01f)
![{\displaystyle a^{0}=1,a^{1}={\frac {1}{2}}(-1+i{\sqrt[{3}]{3}}),a^{2}={\frac {1}{2}}(-1-i{\sqrt[{3}]{3}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ca3baa450e765cb4f8bb2f990d8a395837d1af1)
![{\displaystyle y_{1}={\sqrt[{3}]{t}}_{1}+{\sqrt[{3}]{t}}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e46d19c9af5f0e1c92ff0a91f8b637180d632f2c)
![{\displaystyle y_{2}=a{\sqrt[{3}]{t}}_{1}+a{\sqrt[{3}]{t}}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/469484cbeaf5f56f3924aacb22db93eeaeabccbf)
![{\displaystyle y_{3}=a^{2}{\sqrt[{3}]{t}}_{1}+{\sqrt[{3}]{t}}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6321eb187df7593e5ae90a16e3ffafd110b2abc2)
![{\displaystyle y_{2}=(a-a^{2}){\sqrt[{3}]{\sqrt {\frac {-64}{27}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abaa9562650b193691241066217f74e4d60aa2b3)
![{\displaystyle y_{3}=(a^{2}-a){\sqrt[{3}]{\sqrt {\frac {-64}{27}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b54a23345bcd58c9ab36b84212e8c58e89acd05)
![{\displaystyle y_{2}=(i{\sqrt[{3}]{3}}){\frac {2i}{\sqrt[{3}]{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b7707004262d4eafa6a9b6217e2e20039bbf769)
![{\displaystyle y_{3}=(-i{\sqrt[{3}]{3}}){\frac {2i}{\sqrt[{3}]{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03dd5c16897b8a818849ad68e5dbca2d9bc8c54c)
![{\displaystyle {\begin{aligned}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+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-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-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+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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef8fdcb363d20258b4ec96f33fc248aae912c9ff)