Kubinės lygties sprendimas Kordano metodu
keisti
Duota kubinė lygtis:
y
3
+
a
y
2
+
b
y
+
c
=
0.
{\displaystyle y^{3}+ay^{2}+by+c=0.}
Pakeičiame
y
=
x
−
a
3
,
{\displaystyle y=x-{\frac {a}{3}},}
(
x
−
a
3
)
3
+
a
(
x
−
a
3
)
2
+
b
(
x
−
a
3
)
+
c
=
0
,
{\displaystyle \left(x-{\frac {a}{3}}\right)^{3}+a\left(x-{\frac {a}{3}}\right)^{2}+b\left(x-{\frac {a}{3}}\right)+c=0,}
(
x
3
−
3
⋅
x
2
⋅
a
3
+
3
⋅
x
⋅
(
a
3
)
2
−
(
a
3
)
3
)
+
a
(
x
2
−
2
⋅
x
⋅
a
3
+
(
a
3
)
2
)
+
b
(
x
−
a
3
)
+
c
=
0
,
{\displaystyle \left(x^{3}-3\cdot x^{2}\cdot {\frac {a}{3}}+3\cdot x\cdot \left({\frac {a}{3}}\right)^{2}-\left({\frac {a}{3}}\right)^{3}\right)+a\left(x^{2}-2\cdot x\cdot {\frac {a}{3}}+\left({\frac {a}{3}}\right)^{2}\right)+b\left(x-{\frac {a}{3}}\right)+c=0,}
(
x
3
−
a
x
2
+
3
⋅
x
⋅
a
2
9
−
a
3
27
)
+
a
(
x
2
−
2
a
x
3
+
a
2
9
)
+
b
x
−
a
b
3
+
c
=
0
,
{\displaystyle \left(x^{3}-ax^{2}+3\cdot x\cdot {\frac {a^{2}}{9}}-{\frac {a^{3}}{27}}\right)+a\left(x^{2}-{\frac {2ax}{3}}+{\frac {a^{2}}{9}}\right)+bx-{\frac {ab}{3}}+c=0,}
(
x
3
−
a
x
2
+
a
2
x
3
−
a
3
27
)
+
a
x
2
−
2
a
2
x
3
+
a
3
9
+
b
x
−
a
b
3
+
c
=
0
,
{\displaystyle \left(x^{3}-ax^{2}+{\frac {a^{2}x}{3}}-{\frac {a^{3}}{27}}\right)+ax^{2}-{\frac {2a^{2}x}{3}}+{\frac {a^{3}}{9}}+bx-{\frac {ab}{3}}+c=0,}
x
3
−
a
3
27
−
a
2
x
3
+
a
3
9
+
b
x
−
a
b
3
+
c
=
0
,
{\displaystyle x^{3}-{\frac {a^{3}}{27}}-{\frac {a^{2}x}{3}}+{\frac {a^{3}}{9}}+bx-{\frac {ab}{3}}+c=0,}
x
3
−
a
2
x
3
+
b
x
+
a
3
9
−
a
3
27
−
a
b
3
+
c
=
0
,
{\displaystyle x^{3}-{\frac {a^{2}x}{3}}+bx+{\frac {a^{3}}{9}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,}
x
3
+
(
b
−
a
2
3
)
x
+
3
a
3
27
−
a
3
27
−
a
b
3
+
c
=
0
,
{\displaystyle x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {3a^{3}}{27}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,}
x
3
+
(
b
−
a
2
3
)
x
+
2
a
3
27
−
a
b
3
+
c
=
0.
{\displaystyle x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c=0.}
Pažymime
p
=
b
−
a
2
3
,
q
=
2
a
3
27
−
a
b
3
+
c
{\displaystyle p=b-{\frac {a^{2}}{3}},\quad q={\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c}
x
3
+
p
x
+
q
=
0.
{\displaystyle x^{3}+px+q=0.}
Toliau, tariame, kad lygties
x
3
+
p
x
+
q
=
0
{\displaystyle x^{3}+px+q=0}
x
0
{\displaystyle x_{0}}
p yra tos pačios pagalbinės kvadratinės lygties laisvasis narys (konstanta) padalintas iš 3. Ir tokiu budu sudarome naują pagalbinę kvadratinę funkciją ir lygtį:
f
(
u
)
=
u
2
−
x
0
u
−
p
3
;
{\displaystyle f(u)=u^{2}-x_{0}u-{\frac {p}{3}};}
u
2
−
x
0
u
−
p
3
=
0.
{\displaystyle u^{2}-x_{0}u-{\frac {p}{3}}=0.}
Šios kvadatinės lygties šaknys (sprendiniai) yra
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
u
1
+
u
2
=
−
(
−
x
0
)
=
x
0
{\displaystyle u_{1}+u_{2}=-(-x_{0})=x_{0}}
u
1
⋅
u
2
=
−
p
3
;
−
3
u
1
u
2
=
p
{\displaystyle u_{1}\cdot u_{2}=-{\frac {p}{3}};\quad -3u_{1}u_{2}=p}
Į kubinę lygtį
x
3
+
p
x
+
q
=
0
{\displaystyle x^{3}+px+q=0}
p išreikštą per
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
x
0
{\displaystyle x_{0}}
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
q . Taigi:
x
0
3
+
p
x
0
+
q
=
0
,
{\displaystyle x_{0}^{3}+px_{0}+q=0,}
(
u
1
+
u
2
)
3
−
3
u
1
u
2
(
u
1
+
u
2
)
+
q
=
0
,
{\displaystyle (u_{1}+u_{2})^{3}-3u_{1}u_{2}(u_{1}+u_{2})+q=0,}
u
1
3
+
3
u
1
2
u
2
+
3
u
1
u
2
2
+
u
2
3
−
3
u
1
u
2
(
u
1
+
u
2
)
+
q
=
0
,
{\displaystyle u_{1}^{3}+3u_{1}^{2}u_{2}+3u_{1}u_{2}^{2}+u_{2}^{3}-3u_{1}u_{2}(u_{1}+u_{2})+q=0,}
u
1
3
+
3
u
1
2
u
2
+
3
u
1
u
2
2
+
u
2
3
−
3
u
1
2
u
2
−
3
u
1
u
2
2
+
q
=
0
,
{\displaystyle u_{1}^{3}+3u_{1}^{2}u_{2}+3u_{1}u_{2}^{2}+u_{2}^{3}-3u_{1}^{2}u_{2}-3u_{1}u_{2}^{2}+q=0,}
u
1
3
+
u
2
3
+
q
=
0
,
{\displaystyle u_{1}^{3}+u_{2}^{3}+q=0,}
u
1
3
+
u
2
3
=
−
q
.
{\displaystyle u_{1}^{3}+u_{2}^{3}=-q.}
Vadinasi,
u
1
3
=
z
1
{\displaystyle u_{1}^{3}=z_{1}}
u
2
3
=
z
2
{\displaystyle u_{2}^{3}=z_{2}}
z
2
+
q
z
−
p
3
27
=
0
,
{\displaystyle z^{2}+qz-{\frac {p^{3}}{27}}=0,}
z
1
⋅
z
2
=
u
1
3
⋅
u
2
3
=
(
−
p
3
)
3
=
−
p
3
27
{\displaystyle z_{1}\cdot z_{2}=u_{1}^{3}\cdot u_{2}^{3}=\left(-{\frac {p}{3}}\right)^{3}=-{\frac {p^{3}}{27}}}
z
1
+
z
2
=
u
1
3
+
u
2
3
=
−
q
{\displaystyle z_{1}+z_{2}=u_{1}^{3}+u_{2}^{3}=-q}
Išsprendžiame šią (antrą) kvadratinę lygtį:
z
2
+
q
z
−
p
3
27
=
0
,
{\displaystyle z^{2}+qz-{\frac {p^{3}}{27}}=0,}
(
z
+
q
2
)
2
−
q
2
4
−
p
3
27
=
0
,
{\displaystyle \left(z+{\frac {q}{2}}\right)^{2}-{\frac {q^{2}}{4}}-{\frac {p^{3}}{27}}=0,}
(
z
+
q
2
)
2
=
q
2
4
+
p
3
27
,
{\displaystyle \left(z+{\frac {q}{2}}\right)^{2}={\frac {q^{2}}{4}}+{\frac {p^{3}}{27}},}
z
+
q
2
=
±
q
2
4
+
p
3
27
,
{\displaystyle z+{\frac {q}{2}}=\pm {\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}},}
z
=
−
q
2
±
q
2
4
+
p
3
27
.
{\displaystyle z=-{\frac {q}{2}}\pm {\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}.}
Vadinasi lygties
z
2
+
q
z
−
p
3
27
=
0
{\displaystyle z^{2}+qz-{\frac {p^{3}}{27}}=0}
z
1
=
−
q
2
+
q
2
4
+
p
3
27
;
{\displaystyle z_{1}=-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}};}
z
2
=
−
q
2
−
q
2
4
+
p
3
27
;
{\displaystyle z_{2}=-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}};}
Na, o [pirmos] kvadratinės lygties
u
2
−
x
0
u
−
p
3
=
0
{\displaystyle u^{2}-x_{0}u-{\frac {p}{3}}=0}
z
1
⋅
z
2
=
−
p
3
27
,
{\displaystyle z_{1}\cdot z_{2}=-{\frac {p^{3}}{27}},}
u
1
3
⋅
u
2
3
=
(
−
p
3
)
3
{\displaystyle u_{1}^{3}\cdot u_{2}^{3}=\left(-{\frac {p}{3}}\right)^{3}}
u
1
⋅
u
2
=
−
p
3
{\displaystyle u_{1}\cdot u_{2}=-{\frac {p}{3}}}
u
1
=
z
1
3
=
−
q
2
+
q
2
4
+
p
3
27
3
;
{\displaystyle u_{1}={\sqrt[{3}]{z_{1}}}={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}};}
u
2
=
z
2
3
=
−
q
2
−
q
2
4
+
p
3
27
3
.
{\displaystyle u_{2}={\sqrt[{3}]{z_{2}}}={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}.}
Prisimindami, kad lygties
x
3
+
p
x
+
q
=
0
{\displaystyle x^{3}+px+q=0}
x
0
=
u
1
+
u
2
{\displaystyle x_{0}=u_{1}+u_{2}}
x
0
=
x
1
=
u
1
+
u
2
=
−
q
2
+
q
2
4
+
p
3
27
3
+
−
q
2
−
q
2
4
+
p
3
27
3
.
{\displaystyle x_{0}=x_{1}=u_{1}+u_{2}={\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}}}}}}.}
Kitos dvi kompleksinės šaknys, tenkina lygybę
(
u
1
ϵ
)
⋅
(
u
2
ϵ
2
)
=
−
p
3
{\displaystyle (u_{1}\epsilon )\cdot (u_{2}\epsilon ^{2})=-{\frac {p}{3}}}
(
u
1
ϵ
2
)
⋅
(
u
2
ϵ
)
=
−
p
3
.
{\displaystyle (u_{1}\epsilon ^{2})\cdot (u_{2}\epsilon )=-{\frac {p}{3}}.}
ϵ
=
−
1
2
+
i
3
2
;
ϵ
2
=
−
1
2
−
i
3
2
;
ϵ
3
=
1
,
{\displaystyle \epsilon =-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}};\quad \epsilon ^{2}=-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}};\quad \epsilon ^{3}=1,}
nes
ϵ
2
=
(
−
1
2
+
i
3
2
)
⋅
(
−
1
2
+
i
3
2
)
=
1
4
−
i
3
4
−
i
3
4
+
i
⋅
i
⋅
3
4
=
1
4
−
i
2
3
4
−
3
4
=
−
2
4
−
i
3
2
=
−
1
2
−
i
3
2
.
{\displaystyle \epsilon ^{2}=\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{4}}-i{\frac {\sqrt {3}}{4}}-i{\frac {\sqrt {3}}{4}}+i\cdot i\cdot {\frac {3}{4}}={\frac {1}{4}}-i{\frac {2{\sqrt {3}}}{4}}-{\frac {3}{4}}=-{\frac {2}{4}}-i{\frac {\sqrt {3}}{2}}=-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}.}
ϵ
3
=
ϵ
2
⋅
ϵ
=
(
−
1
2
−
i
3
2
)
⋅
(
−
1
2
+
i
3
2
)
=
1
4
−
i
3
4
+
i
3
4
−
i
⋅
i
⋅
3
4
=
1
4
−
(
−
1
)
⋅
3
4
=
1
4
+
3
4
=
1.
{\displaystyle \epsilon ^{3}=\epsilon ^{2}\cdot \epsilon =\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{4}}-i{\frac {\sqrt {3}}{4}}+i{\frac {\sqrt {3}}{4}}-i\cdot i\cdot {\frac {3}{4}}={\frac {1}{4}}-(-1)\cdot {\frac {3}{4}}={\frac {1}{4}}+{\frac {3}{4}}=1.}
Vadinasi, kitos dvi lygties
x
3
+
p
x
+
q
=
0
{\displaystyle x^{3}+px+q=0}
x
2
=
u
1
ϵ
+
u
2
ϵ
2
=
u
1
(
−
1
2
+
i
3
2
)
+
u
2
(
−
1
2
−
i
3
2
)
=
−
u
1
+
u
2
2
+
i
3
u
1
−
u
2
2
;
{\displaystyle x_{2}=u_{1}\epsilon +u_{2}\epsilon ^{2}=u_{1}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)+u_{2}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)=-{\frac {u_{1}+u_{2}}{2}}+i{\sqrt {3}}{\frac {u_{1}-u_{2}}{2}};}
x
3
=
u
1
ϵ
2
+
u
2
ϵ
=
u
1
(
−
1
2
−
i
3
2
)
+
u
2
(
−
1
2
+
i
3
2
)
=
−
u
1
+
u
2
2
−
i
3
u
1
−
u
2
2
.
{\displaystyle x_{3}=u_{1}\epsilon ^{2}+u_{2}\epsilon =u_{1}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)+u_{2}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)=-{\frac {u_{1}+u_{2}}{2}}-i{\sqrt {3}}{\frac {u_{1}-u_{2}}{2}}.}
Jei sprendžiant lygtį
u
2
−
x
0
u
−
p
3
=
0
{\displaystyle u^{2}-x_{0}u-{\frac {p}{3}}=0}
u
1
=
u
2
{\displaystyle u_{1}=u_{2}}
x
2
=
x
3
=
−
1
2
(
u
1
+
u
2
)
=
−
1
2
−
q
2
+
q
2
4
+
p
3
27
3
−
1
2
−
q
2
−
q
2
4
+
p
3
27
3
=
−
1
2
(
−
q
2
3
+
−
q
2
3
)
=
−
−
q
2
3
=
q
2
3
.
{\displaystyle x_{2}=x_{3}=-{\frac {1}{2}}(u_{1}+u_{2})=-{\frac {1}{2}}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}-{\frac {1}{2}}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}=-{\frac {1}{2}}\left({\sqrt[{3}]{-{\frac {q}{2}}}}+{\sqrt[{3}]{-{\frac {q}{2}}}}\right)=-{\sqrt[{3}]{-{\frac {q}{2}}}}={\sqrt[{3}]{\frac {q}{2}}}.}
Nes tada
q
2
4
+
p
3
27
=
0.
{\displaystyle {\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}=0.}
Iš sąlygos
Δ
3
=
q
2
4
+
p
3
27
=
0
(
20
)
{\displaystyle \Delta _{3}={\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}=0\quad (20)}
išeina, kad
(
−
q
2
)
2
=
(
−
p
3
)
3
.
(
21
)
{\displaystyle (-{\frac {q}{2}})^{2}=(-{\frac {p}{3}})^{3}.\quad (21)}
Jei
p
≠
0
,
{\displaystyle p\neq 0,}
−
q
2
3
=
−
q
2
(
−
q
2
)
2
3
=
−
q
2
(
−
p
3
)
3
3
=
−
q
2
−
p
3
=
3
q
2
p
.
(
22
)
{\displaystyle {\sqrt[{3}]{-{\frac {q}{2}}}}={\frac {-{\frac {q}{2}}}{\sqrt[{3}]{(-{\frac {q}{2}})^{2}}}}={\frac {-{\frac {q}{2}}}{\sqrt[{3}]{(-{\frac {p}{3}})^{3}}}}={\frac {-{\frac {q}{2}}}{-{\frac {p}{3}}}}={\frac {3q}{2p}}.\quad (22)}
Vadinasi, bent viena šaknies reikšmė racionaliai išsireiškia koeficientais p ir q . Parinkę
u
1
=
u
2
=
3
q
2
p
,
{\displaystyle u_{1}=u_{2}={\frac {3q}{2p}},}
u
1
u
2
=
3
q
2
p
⋅
3
q
2
p
=
9
q
2
4
p
2
=
9
p
2
(
−
q
2
)
2
=
9
p
2
⋅
−
p
3
27
=
−
p
3
.
{\displaystyle u_{1}u_{2}={\frac {3q}{2p}}\cdot {\frac {3q}{2p}}={\frac {9q^{2}}{4p^{2}}}={\frac {9}{p^{2}}}(-{\frac {q}{2}})^{2}={\frac {9}{p^{2}}}\cdot {\frac {-p^{3}}{27}}=-{\frac {p}{3}}.}
Iš to turime, kad kai
u
1
=
u
2
{\displaystyle u_{1}=u_{2}}
x
1
=
u
1
+
u
2
=
−
q
2
3
+
−
q
2
3
=
2
−
q
2
3
=
2
⋅
3
q
2
p
=
3
q
p
;
(
22.1
)
{\displaystyle x_{1}=u_{1}+u_{2}={\sqrt[{3}]{-{\frac {q}{2}}}}+{\sqrt[{3}]{-{\frac {q}{2}}}}=2{\sqrt[{3}]{-{\frac {q}{2}}}}=2\cdot {\frac {3q}{2p}}={\frac {3q}{p}};\quad (22.1)}
x
2
=
x
3
=
−
1
2
(
u
1
+
u
2
)
=
−
1
2
(
−
q
2
3
+
−
q
2
3
)
=
−
(
−
q
2
3
)
=
−
3
q
2
p
.
(
22.2
)
{\displaystyle x_{2}=x_{3}=-{\frac {1}{2}}(u_{1}+u_{2})=-{\frac {1}{2}}\left({\sqrt[{3}]{-{\frac {q}{2}}}}+{\sqrt[{3}]{-{\frac {q}{2}}}}\right)=-\left({\sqrt[{3}]{-{\frac {q}{2}}}}\right)=-{\frac {3q}{2p}}.\quad (22.2)}
Raskime lygties
x
3
−
3
x
+
2
=
0
{\displaystyle x^{3}-3x+2=0}
x
1
=
u
1
+
u
2
=
−
q
2
+
q
2
4
+
p
3
27
3
+
−
q
2
−
q
2
4
+
p
3
27
3
=
−
2
2
+
2
2
4
+
(
−
3
)
3
27
3
+
−
2
2
−
2
2
4
+
(
−
3
)
3
27
3
=
{\displaystyle x_{1}=u_{1}+u_{2}={\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}}}}}}={\sqrt[{3}]{-{\frac {2}{2}}+{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {2}{2}}-{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}=}
=
−
1
+
4
4
−
27
27
3
+
−
1
−
4
4
−
27
27
3
=
−
1
+
0
3
+
−
1
−
0
3
=
−
1
3
+
−
1
3
=
−
1
−
1
=
−
2
;
{\displaystyle ={\sqrt[{3}]{-1+{\sqrt {{\frac {4}{4}}-{\frac {27}{27}}}}}}+{\sqrt[{3}]{-1-{\sqrt {{\frac {4}{4}}-{\frac {27}{27}}}}}}={\sqrt[{3}]{-1+{\sqrt {0}}}}+{\sqrt[{3}]{-1-{\sqrt {0}}}}={\sqrt[{3}]{-1}}+{\sqrt[{3}]{-1}}=-1-1=-2;}
x
2
=
u
1
ϵ
+
u
2
ϵ
2
=
u
1
(
−
1
2
+
i
3
2
)
+
u
2
(
−
1
2
−
i
3
2
)
=
{\displaystyle x_{2}=u_{1}\epsilon +u_{2}\epsilon ^{2}=u_{1}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)+u_{2}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)=}
=
(
−
1
2
+
i
3
2
)
−
q
2
+
q
2
4
+
p
3
27
3
+
(
−
1
2
−
i
3
2
)
−
q
2
−
q
2
4
+
p
3
27
3
=
{\displaystyle =\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}=}
=
(
−
1
2
+
i
3
2
)
−
2
2
+
2
2
4
+
(
−
3
)
3
27
3
+
(
−
1
2
−
i
3
2
)
−
2
2
−
2
2
4
+
(
−
3
)
3
27
3
=
(
−
1
2
+
i
3
2
)
−
1
3
+
(
−
1
2
−
i
3
2
)
−
1
3
=
{\displaystyle =\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}+{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}+\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}-{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}=\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}+\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}=}
=
−
(
−
1
2
+
i
3
2
)
−
(
−
1
2
−
i
3
2
)
=
1
2
−
i
3
2
+
1
2
+
i
3
2
=
1
2
+
1
2
=
1.
{\displaystyle =-\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)-\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}+{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}={\frac {1}{2}}+{\frac {1}{2}}=1.}
x
3
=
u
1
ϵ
2
+
u
2
ϵ
=
u
1
(
−
1
2
−
i
3
2
)
+
u
2
(
−
1
2
+
i
3
2
)
=
{\displaystyle x_{3}=u_{1}\epsilon ^{2}+u_{2}\epsilon =u_{1}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)+u_{2}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)=}
=
(
−
1
2
−
i
3
2
)
−
q
2
+
q
2
4
+
p
3
27
3
+
(
−
1
2
+
i
3
2
)
−
q
2
−
q
2
4
+
p
3
27
3
=
{\displaystyle =\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}=}
=
(
−
1
2
−
i
3
2
)
−
2
2
+
2
2
4
+
(
−
3
)
3
27
3
+
(
−
1
2
+
i
3
2
)
−
2
2
−
2
2
4
+
(
−
3
)
3
27
3
=
(
−
1
2
−
i
3
2
)
−
1
3
+
(
−
1
2
+
i
3
2
)
−
1
3
=
{\displaystyle =\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}+{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}+\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}-{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}=\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}+\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}=}
=
−
(
−
1
2
−
i
3
2
)
−
(
−
1
2
+
i
3
2
)
=
1
2
+
i
3
2
+
1
2
−
i
3
2
=
1
2
+
1
2
=
1.
{\displaystyle =-\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)-\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}+{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}={\frac {1}{2}}+{\frac {1}{2}}=1.}
Įstatę į lygtį
x
3
−
3
x
+
2
=
0
{\displaystyle x^{3}-3x+2=0}
x
1
=
−
2
{\displaystyle x_{1}=-2}
(
−
2
)
3
−
3
⋅
(
−
2
)
+
2
=
−
8
+
6
+
2
=
0.
{\displaystyle (-2)^{3}-3\cdot (-2)+2=-8+6+2=0.}
Įstatę į lygtį
x
3
−
3
x
+
2
=
0
{\displaystyle x^{3}-3x+2=0}
x
2
=
1
{\displaystyle x_{2}=1}
x
3
=
1
{\displaystyle x_{3}=1}
1
3
−
3
⋅
1
+
2
=
1
−
3
+
2
=
0.
{\displaystyle 1^{3}-3\cdot 1+2=1-3+2=0.}
Greičiau visus sprendinius randame iš formulių (22.1) ir (22.2):
x
1
=
3
q
p
=
3
⋅
2
−
3
=
−
2
;
{\displaystyle x_{1}={\frac {3q}{p}}={\frac {3\cdot 2}{-3}}=-2;}
x
2
=
x
3
=
−
3
q
2
p
=
−
3
⋅
2
2
⋅
(
−
3
)
=
1.
{\displaystyle x_{2}=x_{3}=-{\frac {3q}{2p}}=-{\frac {3\cdot 2}{2\cdot (-3)}}=1.}
Kitoks kubinės lygties sprendimo būdas
keisti
Duota pilna kubinė lygtis:
a
x
3
+
b
x
2
+
c
x
+
d
=
0.
{\displaystyle ax^{3}+bx^{2}+cx+d=0.}
Eliminuojame
b
x
2
{\displaystyle bx^{2}}
x
=
y
−
b
3
a
.
{\displaystyle x=y-{\frac {b}{3a}}.}
a
(
y
−
b
3
a
)
3
+
b
(
y
−
b
3
a
)
2
+
c
(
y
−
b
3
a
)
+
d
=
0
,
{\displaystyle a(y-{\frac {b}{3a}})^{3}+b(y-{\frac {b}{3a}})^{2}+c(y-{\frac {b}{3a}})+d=0,}
a
(
y
3
−
3
y
2
b
3
a
+
3
y
(
b
3
a
)
2
−
(
b
3
a
)
3
)
+
b
(
y
2
−
2
y
b
3
a
+
(
b
3
a
)
2
)
+
c
(
y
−
b
3
a
)
+
d
=
0
,
{\displaystyle a(y^{3}-3y^{2}{\frac {b}{3a}}+3y({\frac {b}{3a}})^{2}-({\frac {b}{3a}})^{3})+b(y^{2}-2y{\frac {b}{3a}}+({\frac {b}{3a}})^{2})+c(y-{\frac {b}{3a}})+d=0,}
(
y
3
−
y
2
b
a
+
y
b
2
a
2
−
b
3
27
a
3
)
+
b
a
(
y
2
−
y
2
b
3
a
+
b
2
9
a
2
)
+
c
a
(
y
−
b
3
a
)
+
d
a
=
0
,
{\displaystyle (y^{3}-y^{2}{\frac {b}{a}}+y{\frac {b^{2}}{a^{2}}}-{\frac {b^{3}}{27a^{3}}})+{\frac {b}{a}}(y^{2}-y{\frac {2b}{3a}}+{\frac {b^{2}}{9a^{2}}})+{\frac {c}{a}}(y-{\frac {b}{3a}})+{\frac {d}{a}}=0,}
(
y
3
−
y
2
b
a
+
y
b
2
a
2
−
b
3
27
a
3
)
+
(
y
2
b
a
−
y
2
b
2
3
a
2
+
b
3
9
a
3
)
+
(
y
c
a
−
b
c
3
a
2
)
+
d
a
=
0
,
{\displaystyle (y^{3}-y^{2}{\frac {b}{a}}+y{\frac {b^{2}}{a^{2}}}-{\frac {b^{3}}{27a^{3}}})+(y^{2}{\frac {b}{a}}-y{\frac {2b^{2}}{3a^{2}}}+{\frac {b^{3}}{9a^{3}}})+(y{\frac {c}{a}}-{\frac {bc}{3a^{2}}})+{\frac {d}{a}}=0,}
y
3
+
y
2
(
−
b
a
+
b
a
)
+
y
(
b
2
a
2
−
2
b
2
3
a
2
+
c
a
)
−
b
3
27
a
3
+
b
3
9
a
3
−
b
c
3
a
2
+
d
a
=
0
,
{\displaystyle y^{3}+y^{2}(-{\frac {b}{a}}+{\frac {b}{a}})+y({\frac {b^{2}}{a^{2}}}-{\frac {2b^{2}}{3a^{2}}}+{\frac {c}{a}})-{\frac {b^{3}}{27a^{3}}}+{\frac {b^{3}}{9a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}=0,}
y
3
+
(
c
a
−
b
2
3
a
2
)
y
+
(
d
a
+
2
b
3
27
a
3
−
b
c
3
a
2
)
=
0.
{\displaystyle y^{3}+({\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}})y+({\frac {d}{a}}+{\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}})=0.}
Pažymime:
V
=
c
a
−
b
2
3
a
2
;
{\displaystyle V={\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}};}
P
=
d
a
+
2
b
3
27
a
3
−
b
c
3
a
2
.
{\displaystyle P={\frac {d}{a}}+{\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}.}
Turime kubinę lygtį:
y
3
+
V
y
+
P
=
0.
{\displaystyle y^{3}+Vy+P=0.}
Parenkame, kad
V
=
3
s
t
{\displaystyle V=3st}
y
=
s
−
t
{\displaystyle y=s-t}
y
3
+
3
s
t
y
+
P
=
0
,
{\displaystyle y^{3}+3sty+P=0,}
(
s
−
t
)
3
+
3
s
t
(
s
−
t
)
+
P
=
0
,
{\displaystyle (s-t)^{3}+3st(s-t)+P=0,}
s
3
−
3
s
2
t
+
3
s
t
2
−
t
3
+
3
s
t
(
s
−
t
)
+
P
=
0
,
{\displaystyle s^{3}-3s^{2}t+3st^{2}-t^{3}+3st(s-t)+P=0,}
s
3
−
3
s
2
t
+
3
s
t
2
−
t
3
+
3
s
2
t
−
3
s
t
2
+
P
=
0
,
{\displaystyle s^{3}-3s^{2}t+3st^{2}-t^{3}+3s^{2}t-3st^{2}+P=0,}
s
3
−
t
3
+
P
=
0
,
{\displaystyle s^{3}-t^{3}+P=0,}
P
=
t
3
−
s
3
.
{\displaystyle P=t^{3}-s^{3}.}
Iš lygybės
V
=
3
s
t
{\displaystyle V=3st}
t
=
V
3
s
.
{\displaystyle t={\frac {V}{3s}}.}
P
=
(
V
3
s
)
3
−
s
3
,
{\displaystyle P=\left({\frac {V}{3s}}\right)^{3}-s^{3},}
V
3
27
s
3
−
s
3
=
P
,
{\displaystyle {\frac {V^{3}}{27s^{3}}}-s^{3}=P,}
V
3
−
27
s
6
=
27
s
3
P
,
{\displaystyle V^{3}-27s^{6}=27s^{3}P,}
V
3
−
27
s
6
−
27
s
3
P
=
0
,
{\displaystyle V^{3}-27s^{6}-27s^{3}P=0,}
27
s
6
+
27
s
3
P
−
V
3
=
0.
{\displaystyle 27s^{6}+27s^{3}P-V^{3}=0.}
Sprendžiame kaip kvadratinę lygtį, radę diskriminantą:
D
=
(
27
P
)
2
−
4
⋅
27
⋅
(
−
V
3
)
=
729
P
2
+
108
V
3
;
{\displaystyle D=(27P)^{2}-4\cdot 27\cdot (-V^{3})=729P^{2}+108V^{3};}
s
3
=
−
27
P
2
⋅
27
±
27
2
P
2
+
4
⋅
27
V
3
2
⋅
27
,
{\displaystyle s^{3}={\frac {-27P}{2\cdot 27}}\pm {\frac {\sqrt {27^{2}P^{2}+4\cdot 27V^{3}}}{2\cdot 27}},}
s
3
=
−
P
2
±
P
2
+
4
V
3
27
2
,
{\displaystyle s^{3}={\frac {-P}{2}}\pm {\frac {\sqrt {P^{2}+{\frac {4V^{3}}{27}}}}{2}},}
s
3
=
−
P
2
±
P
2
4
+
V
3
27
.
{\displaystyle s^{3}={\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}.}
s
=
−
P
2
±
P
2
4
+
V
3
27
3
.
{\displaystyle s={\sqrt[{3}]{{\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}
Prisimename, kad:
P
=
t
3
−
s
3
,
{\displaystyle P=t^{3}-s^{3},}
P
=
t
3
−
(
−
P
2
±
P
2
4
+
V
3
27
)
,
{\displaystyle P=t^{3}-({\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}),}
P
=
t
3
+
P
2
∓
P
2
4
+
V
3
27
,
{\displaystyle P=t^{3}+{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}},}
−
t
3
=
−
P
+
P
2
∓
P
2
4
+
V
3
27
,
{\displaystyle -t^{3}=-P+{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}},}
−
t
3
=
−
P
2
∓
P
2
4
+
V
3
27
,
{\displaystyle -t^{3}=-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}},}
t
3
=
P
2
±
P
2
4
+
V
3
27
.
{\displaystyle t^{3}={\frac {P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}.}
t
=
P
2
±
P
2
4
+
V
3
27
3
.
{\displaystyle t={\sqrt[{3}]{{\frac {P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}
Prisimename, kad
y
=
s
−
t
{\displaystyle y=s-t}
y
=
s
−
t
=
−
P
2
±
P
2
4
+
V
3
27
3
−
P
2
±
P
2
4
+
V
3
27
3
=
−
P
2
±
P
2
4
+
V
3
27
3
+
−
P
2
∓
P
2
4
+
V
3
27
3
.
{\displaystyle y=s-t={\sqrt[{3}]{{\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}-{\sqrt[{3}]{{\frac {P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}
Tai
y
1
=
−
P
2
+
P
2
4
+
V
3
27
3
+
−
P
2
−
P
2
4
+
V
3
27
3
{\displaystyle y_{1}={\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}}
y
2
=
−
P
2
−
P
2
4
+
V
3
27
3
+
−
P
2
+
P
2
4
+
V
3
27
3
{\displaystyle y_{2}={\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}}
y
1
=
y
2
.
{\displaystyle y_{1}=y_{2}.}
Randame lygties
a
x
3
+
b
x
2
+
c
x
+
d
{\displaystyle ax^{3}+bx^{2}+cx+d}
x
1
=
y
−
b
3
a
=
−
P
2
+
P
2
4
+
V
3
27
3
+
−
P
2
−
P
2
4
+
V
3
27
3
−
b
3
a
,
{\displaystyle x_{1}=y-{\frac {b}{3a}}={\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}-{\frac {b}{3a}},}
čia
V
=
c
a
−
b
2
3
a
2
,
{\displaystyle V={\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}},}
P
=
d
a
+
2
b
3
27
a
3
−
b
c
3
a
2
.
{\displaystyle P={\frac {d}{a}}+{\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}.}
Kitoks kubinės lygties sprendimo būdas (2)
keisti
Duota pilna kubinė lygtis:
a
x
3
+
b
x
2
+
c
x
+
d
=
0.
{\displaystyle ax^{3}+bx^{2}+cx+d=0.}
Eliminuojame
b
x
2
{\displaystyle bx^{2}}
x
=
y
−
b
3
a
.
{\displaystyle x=y-{\frac {b}{3a}}.}
a
(
y
−
b
3
a
)
3
+
b
(
y
−
b
3
a
)
2
+
c
(
y
−
b
3
a
)
+
d
=
0
,
{\displaystyle a(y-{\frac {b}{3a}})^{3}+b(y-{\frac {b}{3a}})^{2}+c(y-{\frac {b}{3a}})+d=0,}
a
(
y
3
−
3
y
2
b
3
a
+
3
y
(
b
3
a
)
2
−
(
b
3
a
)
3
)
+
b
(
y
2
−
2
y
b
3
a
+
(
b
3
a
)
2
)
+
c
(
y
−
b
3
a
)
+
d
=
0
,
{\displaystyle a(y^{3}-3y^{2}{\frac {b}{3a}}+3y({\frac {b}{3a}})^{2}-({\frac {b}{3a}})^{3})+b(y^{2}-2y{\frac {b}{3a}}+({\frac {b}{3a}})^{2})+c(y-{\frac {b}{3a}})+d=0,}
(
y
3
−
y
2
b
a
+
y
b
2
a
2
−
b
3
27
a
3
)
+
b
a
(
y
2
−
y
2
b
3
a
+
b
2
9
a
2
)
+
c
a
(
y
−
b
3
a
)
+
d
a
=
0
,
{\displaystyle (y^{3}-y^{2}{\frac {b}{a}}+y{\frac {b^{2}}{a^{2}}}-{\frac {b^{3}}{27a^{3}}})+{\frac {b}{a}}(y^{2}-y{\frac {2b}{3a}}+{\frac {b^{2}}{9a^{2}}})+{\frac {c}{a}}(y-{\frac {b}{3a}})+{\frac {d}{a}}=0,}
(
y
3
−
y
2
b
a
+
y
b
2
a
2
−
b
3
27
a
3
)
+
(
y
2
b
a
−
y
2
b
2
3
a
2
+
b
3
9
a
3
)
+
(
y
c
a
−
b
c
3
a
2
)
+
d
a
=
0
,
{\displaystyle (y^{3}-y^{2}{\frac {b}{a}}+y{\frac {b^{2}}{a^{2}}}-{\frac {b^{3}}{27a^{3}}})+(y^{2}{\frac {b}{a}}-y{\frac {2b^{2}}{3a^{2}}}+{\frac {b^{3}}{9a^{3}}})+(y{\frac {c}{a}}-{\frac {bc}{3a^{2}}})+{\frac {d}{a}}=0,}
y
3
+
y
2
(
−
b
a
+
b
a
)
+
y
(
b
2
a
2
−
2
b
2
3
a
2
+
c
a
)
−
b
3
27
a
3
+
b
3
9
a
3
−
b
c
3
a
2
+
d
a
=
0
,
{\displaystyle y^{3}+y^{2}(-{\frac {b}{a}}+{\frac {b}{a}})+y({\frac {b^{2}}{a^{2}}}-{\frac {2b^{2}}{3a^{2}}}+{\frac {c}{a}})-{\frac {b^{3}}{27a^{3}}}+{\frac {b^{3}}{9a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}=0,}
y
3
+
(
c
a
−
b
2
3
a
2
)
y
+
(
d
a
+
2
b
3
27
a
3
−
b
c
3
a
2
)
=
0.
{\displaystyle y^{3}+({\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}})y+({\frac {d}{a}}+{\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}})=0.}
Pažymime:
V
=
c
a
−
b
2
3
a
2
;
{\displaystyle V={\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}};}
P
=
d
a
+
2
b
3
27
a
3
−
b
c
3
a
2
.
{\displaystyle P={\frac {d}{a}}+{\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}.}
Turime kubinę lygtį:
y
3
+
V
y
+
P
=
0.
{\displaystyle y^{3}+Vy+P=0.}
Parenkame, kad
V
=
−
3
s
t
{\displaystyle V=-3st}
y
=
s
+
t
{\displaystyle y=s+t}
y
3
−
3
s
t
y
+
P
=
0
,
{\displaystyle y^{3}-3sty+P=0,}
(
s
+
t
)
3
−
3
s
t
(
s
+
t
)
+
P
=
0
,
{\displaystyle (s+t)^{3}-3st(s+t)+P=0,}
s
3
+
3
s
2
t
+
3
s
t
2
+
t
3
−
3
s
t
(
s
+
t
)
+
P
=
0
,
{\displaystyle s^{3}+3s^{2}t+3st^{2}+t^{3}-3st(s+t)+P=0,}
s
3
+
3
s
2
t
+
3
s
t
2
+
t
3
−
3
s
2
t
−
3
s
t
2
+
P
=
0
,
{\displaystyle s^{3}+3s^{2}t+3st^{2}+t^{3}-3s^{2}t-3st^{2}+P=0,}
s
3
+
t
3
+
P
=
0
,
{\displaystyle s^{3}+t^{3}+P=0,}
P
=
−
t
3
−
s
3
.
{\displaystyle P=-t^{3}-s^{3}.}
Iš lygybės
V
=
−
3
s
t
{\displaystyle V=-3st}
t
=
−
V
3
s
.
{\displaystyle t=-{\frac {V}{3s}}.}
P
=
−
(
−
V
3
s
)
3
−
s
3
,
{\displaystyle P=-\left(-{\frac {V}{3s}}\right)^{3}-s^{3},}
V
3
27
s
3
−
s
3
=
P
,
{\displaystyle {\frac {V^{3}}{27s^{3}}}-s^{3}=P,}
V
3
−
27
s
6
=
27
s
3
P
,
{\displaystyle V^{3}-27s^{6}=27s^{3}P,}
V
3
−
27
s
6
−
27
s
3
P
=
0
,
{\displaystyle V^{3}-27s^{6}-27s^{3}P=0,}
27
s
6
+
27
s
3
P
−
V
3
=
0.
{\displaystyle 27s^{6}+27s^{3}P-V^{3}=0.}
Sprendžiame kaip kvadratinę lygtį, radę diskriminantą:
D
=
(
27
P
)
2
−
4
⋅
27
⋅
(
−
V
3
)
=
729
P
2
+
108
V
3
;
{\displaystyle D=(27P)^{2}-4\cdot 27\cdot (-V^{3})=729P^{2}+108V^{3};}
s
3
=
−
27
P
2
⋅
27
±
27
2
P
2
+
4
⋅
27
V
3
2
⋅
27
,
{\displaystyle s^{3}={\frac {-27P}{2\cdot 27}}\pm {\frac {\sqrt {27^{2}P^{2}+4\cdot 27V^{3}}}{2\cdot 27}},}
s
3
=
−
P
2
±
P
2
+
4
V
3
27
2
,
{\displaystyle s^{3}={\frac {-P}{2}}\pm {\frac {\sqrt {P^{2}+{\frac {4V^{3}}{27}}}}{2}},}
s
3
=
−
P
2
±
P
2
4
+
V
3
27
.
{\displaystyle s^{3}={\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}.}
s
=
−
P
2
±
P
2
4
+
V
3
27
3
.
{\displaystyle s={\sqrt[{3}]{{\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}
Prisimename, kad:
P
=
−
t
3
−
s
3
,
{\displaystyle P=-t^{3}-s^{3},}
P
=
−
t
3
−
(
−
P
2
±
P
2
4
+
V
3
27
)
,
{\displaystyle P=-t^{3}-({\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}),}
P
=
−
t
3
+
P
2
∓
P
2
4
+
V
3
27
,
{\displaystyle P=-t^{3}+{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}},}
t
3
=
−
P
+
P
2
∓
P
2
4
+
V
3
27
,
{\displaystyle t^{3}=-P+{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}},}
t
3
=
−
P
2
∓
P
2
4
+
V
3
27
.
{\displaystyle t^{3}=-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}.}
t
=
−
P
2
∓
P
2
4
+
V
3
27
3
.
{\displaystyle t={\sqrt[{3}]{-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}
Prisimename, kad
y
=
s
+
t
{\displaystyle y=s+t}
y
=
s
+
t
=
−
P
2
±
P
2
4
+
V
3
27
3
+
−
P
2
∓
P
2
4
+
V
3
27
3
.
{\displaystyle y=s+t={\sqrt[{3}]{{\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}
Tai
y
1
=
−
P
2
+
P
2
4
+
V
3
27
3
+
−
P
2
−
P
2
4
+
V
3
27
3
{\displaystyle y_{1}={\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}}
y
2
=
−
P
2
−
P
2
4
+
V
3
27
3
+
−
P
2
+
P
2
4
+
V
3
27
3
{\displaystyle y_{2}={\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}}
y
1
=
y
2
.
{\displaystyle y_{1}=y_{2}.}
Randame lygties
a
x
3
+
b
x
2
+
c
x
+
d
{\displaystyle ax^{3}+bx^{2}+cx+d}
x
1
=
y
−
b
3
a
=
−
P
2
+
P
2
4
+
V
3
27
3
+
−
P
2
−
P
2
4
+
V
3
27
3
−
b
3
a
,
{\displaystyle x_{1}=y-{\frac {b}{3a}}={\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}-{\frac {b}{3a}},}
čia
V
=
c
a
−
b
2
3
a
2
,
{\displaystyle V={\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}},}
P
=
d
a
+
2
b
3
27
a
3
−
b
c
3
a
2
.
{\displaystyle P={\frac {d}{a}}+{\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}.}
Lygties sprendimo tikslas yra rasti išreikštą
P
{\displaystyle P}
s ir t (ar V ). Nes jau turime išreikštą V per s ir t (
V
=
−
3
s
t
;
y
=
s
+
t
{\displaystyle V=-3st;\;y=s+t}
P , kai pakelsime
(
s
+
t
)
{\displaystyle (s+t)}
3
s
t
(
s
+
t
)
=
3
s
2
t
+
3
s
t
2
.
{\displaystyle 3st(s+t)=3s^{2}t+3st^{2}.}
P
=
−
s
3
−
t
3
.
{\displaystyle P=-s^{3}-t^{3}.}
(
s
+
t
)
3
−
3
s
t
(
s
+
t
)
+
P
=
0
,
{\displaystyle (s+t)^{3}-3st(s+t)+P=0,}
s
3
+
3
s
2
t
+
3
s
t
2
+
t
3
−
3
s
t
(
s
+
t
)
+
P
=
0
,
{\displaystyle s^{3}+3s^{2}t+3st^{2}+t^{3}-3st(s+t)+P=0,}
s
3
+
t
3
+
P
=
0.
{\displaystyle s^{3}+t^{3}+P=0.}
Tada toliau gana nesunku rasti s ir t iš sistemos
{
V
+
3
s
t
=
0
,
s
3
+
t
3
+
P
=
0
,
{\displaystyle {\begin{cases}V+3st=0,&\\s^{3}+t^{3}+P=0,&\end{cases}}}
o tada ir y , žinant, kad
y
=
s
+
t
.
{\displaystyle y=s+t.}
![{\displaystyle u_{1}={\sqrt[{3}]{z_{1}}}={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/337ef7f8b3132203cce9ec13de634d80111a94b5)
![{\displaystyle u_{2}={\sqrt[{3}]{z_{2}}}={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73cfe271c887adf164a0ca1f7cf25add69ef6468)
![{\displaystyle x_{0}=x_{1}=u_{1}+u_{2}={\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/3abed0ce0cb69aa2527810f6a13c891cf5ece214)
![{\displaystyle x_{2}=x_{3}=-{\frac {1}{2}}(u_{1}+u_{2})=-{\frac {1}{2}}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}-{\frac {1}{2}}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}=-{\frac {1}{2}}\left({\sqrt[{3}]{-{\frac {q}{2}}}}+{\sqrt[{3}]{-{\frac {q}{2}}}}\right)=-{\sqrt[{3}]{-{\frac {q}{2}}}}={\sqrt[{3}]{\frac {q}{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70ca254c82186349fb921667f2dae004a948d36c)
![{\displaystyle {\sqrt[{3}]{-{\frac {q}{2}}}}={\frac {-{\frac {q}{2}}}{\sqrt[{3}]{(-{\frac {q}{2}})^{2}}}}={\frac {-{\frac {q}{2}}}{\sqrt[{3}]{(-{\frac {p}{3}})^{3}}}}={\frac {-{\frac {q}{2}}}{-{\frac {p}{3}}}}={\frac {3q}{2p}}.\quad (22)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfe9d6bcdb01f9bcd4ee1ae1f298a9b99684f4f7)
![{\displaystyle x_{1}=u_{1}+u_{2}={\sqrt[{3}]{-{\frac {q}{2}}}}+{\sqrt[{3}]{-{\frac {q}{2}}}}=2{\sqrt[{3}]{-{\frac {q}{2}}}}=2\cdot {\frac {3q}{2p}}={\frac {3q}{p}};\quad (22.1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d551ccfac834ec6394af697dc66ab9dea65d3ad)
![{\displaystyle x_{2}=x_{3}=-{\frac {1}{2}}(u_{1}+u_{2})=-{\frac {1}{2}}\left({\sqrt[{3}]{-{\frac {q}{2}}}}+{\sqrt[{3}]{-{\frac {q}{2}}}}\right)=-\left({\sqrt[{3}]{-{\frac {q}{2}}}}\right)=-{\frac {3q}{2p}}.\quad (22.2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/799f17da57bdf8137e9a1cde17362a5204cf8525)
![{\displaystyle x_{1}=u_{1}+u_{2}={\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}}}}}}={\sqrt[{3}]{-{\frac {2}{2}}+{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {2}{2}}-{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ab94333523761ab2c2aab992454e73a555e888d)
![{\displaystyle ={\sqrt[{3}]{-1+{\sqrt {{\frac {4}{4}}-{\frac {27}{27}}}}}}+{\sqrt[{3}]{-1-{\sqrt {{\frac {4}{4}}-{\frac {27}{27}}}}}}={\sqrt[{3}]{-1+{\sqrt {0}}}}+{\sqrt[{3}]{-1-{\sqrt {0}}}}={\sqrt[{3}]{-1}}+{\sqrt[{3}]{-1}}=-1-1=-2;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f669c0901eeaee1479f45ad2c7ad1a4161e38eb6)
![{\displaystyle =\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/265d2ad6be691f24c8a467a0b3c6dfae45e0cea1)
![{\displaystyle =\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}+{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}+\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}-{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}=\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}+\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01a12c7113a6664594192c30f0d6470aa2641d46)
![{\displaystyle =\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22e6fdfe3fe006f24bca2f73f64e61165aec7165)
![{\displaystyle =\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}+{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}+\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {2}{2}}-{\sqrt {{\frac {2^{2}}{4}}+{\frac {(-3)^{3}}{27}}}}}}=\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}+\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-1}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/517a2acf23c92b77886a7ec1e0144c5d63029f47)
![{\displaystyle s={\sqrt[{3}]{{\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31ee7c4d3648ca29f85f37b68bebfbdc888b5ac1)
![{\displaystyle t={\sqrt[{3}]{{\frac {P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3bcbedfef2576dc76b9c295c41e4f9cd71809134)
![{\displaystyle y=s-t={\sqrt[{3}]{{\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}-{\sqrt[{3}]{{\frac {P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bbc45c79b395a1bdf1f0366115dbc3db94a2d42)
![{\displaystyle y_{1}={\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c6546d4f40bc48f8b0d96cd60206862892bbf61)
![{\displaystyle y_{2}={\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86ae94f45baa2bb436b603e55cb392bdf9d0be17)
![{\displaystyle x_{1}=y-{\frac {b}{3a}}={\sqrt[{3}]{-{\frac {P}{2}}+{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}-{\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}-{\frac {b}{3a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8107ad79c5ab9a7a717851c793e0477adda00f93)
![{\displaystyle t={\sqrt[{3}]{-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb9f0762892c50bac7da0d88ce22c63ae7290cc4)
![{\displaystyle y=s+t={\sqrt[{3}]{{\frac {-P}{2}}\pm {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {P}{2}}\mp {\sqrt {{\frac {P^{2}}{4}}+{\frac {V^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/003ab466b904ad35a53b5b6df058001f53754313)
