該頁面尚未檢查
元數一而次數三之方程，是謂一元三次方程，或立方函數之方程也。常書一元三次方程為 a x 3 + b x 2 + c x + d = 0 {\displaystyle ax^{3}+bx^{2}+cx+d=0} 。
欲解一元三次方程，必先去二次項，而成簡版方程。後以卡贊諾或韋達之法，皆能得解。其終式如下： x 1 = − b 3 a + − b 3 27 a 3 + b c 6 a 2 − d 2 a + ( − b 3 27 a 3 + b c 6 a 2 − d 2 a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 + − b 3 27 a 3 + b c 6 a 2 − d 2 a − ( − b 3 27 a 3 + b c 6 a 2 − d 2 a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 {\displaystyle x_{1}=-{\frac {b}{3a}}+{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\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}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}
x 2 = − b 3 a + − 1 − i 3 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a + ( − b 3 27 a 3 + b c 6 a 2 − d 2 a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 + − 1 + i 3 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a − ( − b 3 27 a 3 + b c 6 a 2 − d 2 a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 {\displaystyle x_{2}=-{\frac {b}{3a}}+{\frac {-1-i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}+{\frac {-1+i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}
x 3 = − b 3 a + − 1 + i 3 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a + ( − b 3 27 a 3 + b c 6 a 2 − d 2 a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 + − 1 − i 3 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a − ( − b 3 27 a 3 + b c 6 a 2 − d 2 a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 {\displaystyle x_{3}=-{\frac {b}{3a}}+{\frac {-1+i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}+{\frac {-1-i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}
欲求甚解者，必先去其二次項。此乃求解之要步也，以便其思路，易生求解之法於心中。簡化時，全式除以 a {\displaystyle a} ，以 x = t − b 3 a {\displaystyle x=t-{\frac {b}{3a}}} 代之，得 t 3 + p t + q = 0 {\displaystyle t^{3}+pt+q=0} 之制，其中： p = c 3 a − b 2 3 a 2 {\displaystyle p={\frac {c}{3a}}-{\frac {b^{2}}{3a^{2}}}}
q = 2 b 3 27 a 3 − b c 3 a 2 + d a {\displaystyle q={\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}}
證： a x 3 + b x 2 + c x + d = 0 {\displaystyle ax^{3}+bx^{2}+cx+d=0}
x 3 + b a x 2 + c a x + d a = 0 {\displaystyle x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x+{\frac {d}{a}}=0}
( t − b 3 a ) 3 + b a ( t − b 3 a ) 2 + c a ( t − b 3 a ) + d a = 0 {\displaystyle (t-{\frac {b}{3a}})^{3}+{\frac {b}{a}}(t-{\frac {b}{3a}})^{2}+{\frac {c}{a}}(t-{\frac {b}{3a}})+{\frac {d}{a}}=0}
t 3 − b a t 2 + b 2 3 a 2 t − b 3 27 a 3 + b a t 2 − 2 b 2 3 a 2 t + b 3 9 a 3 + c a t − b c 3 a 2 + d a = 0 {\displaystyle t^{3}-{\frac {b}{a}}t^{2}+{\frac {b^{2}}{3a^{2}}}t-{\frac {b^{3}}{27a^{3}}}+{\frac {b}{a}}t^{2}-{\frac {2b^{2}}{3a^{2}}}t+{\frac {b^{3}}{9a^{3}}}+{\frac {c}{a}}t-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}=0}
t 3 + ( c a − b 2 3 a 2 ) t + ( 2 b 3 27 a 3 − b c 3 a 2 + d a ) = 0 {\displaystyle t^{3}+({\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}})t+({\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}})=0}
t 3 + p t + q = 0 {\displaystyle t^{3}+pt+q=0}
終步則以 t {\displaystyle t} 之三可能值加諸 − b 3 a {\displaystyle -{\frac {b}{3a}}} 成解也。
此步后，毋論卡贊諾之法或韋達之法，具解 t {\displaystyle t} 也。
大賢卡贊諾云：設 t = u + v {\displaystyle t=u+v} 。得：
( u + v ) 3 + p ( u + v ) + q = 0 {\displaystyle (u+v)^{3}+p(u+v)+q=0}
u 3 + v 3 + ( 3 u v + p ) ( u + v ) + q = 0 {\displaystyle u^{3}+v^{3}+(3uv+p)(u+v)+q=0}
又：設 3 u v + p = 0 {\displaystyle 3uv+p=0} 。則： { u 3 + v 3 = − q . . . ( 1 ) u 3 v 3 = − p 3 27 . . . ( 2 ) {\displaystyle {\begin{cases}u^{3}+v^{3}=-q...(1)\\u^{3}v^{3}=-{\frac {p^{3}}{27}}...(2)\\\end{cases}}}
u 3 {\displaystyle u^{3}} 和 v 3 {\displaystyle v^{3}} 皆 y 2 + q y − p 3 27 = 0 {\displaystyle y^{2}+qy-{\frac {p^{3}}{27}}=0} 之解。以一元二次方程公式可解之。得：
u 3 , v 3 = − q ± q 2 + 4 27 p 3 2 = − q 2 ± q 2 4 + p 3 27 {\displaystyle u^{3},v^{3}={\frac {-q\pm {\sqrt {q^{2}+{\frac {4}{27}}p^{3}}}}{2}}=-{\frac {q}{2}}\pm {\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}
復以單位根求 u , v {\displaystyle u,v} 之三可能值，以 u + v {\displaystyle u+v} 得 t {\displaystyle t} （實解者相加，複解者系數乃共軛者相加）。遂以 x = t − b 3 a {\displaystyle x=t-{\frac {b}{3a}}} 得解也。
大賢韋達云：設 t = z + k z {\displaystyle t=z+{\frac {k}{z}}} 。得：
( z + k z ) 3 + p ( z + k z ) + q = 0 {\displaystyle (z+{\frac {k}{z}})^{3}+p(z+{\frac {k}{z}})+q=0}
z 3 + ( 3 k + p ) ( z + k z ) + k 3 z 3 + q = 0 {\displaystyle z^{3}+(3k+p)(z+{\frac {k}{z}})+{\frac {k^{3}}{z^{3}}}+q=0}
又：設 3 k + p = 0 {\displaystyle 3k+p=0} 。則： z 3 + q + k 3 z 3 = 0 {\displaystyle z^{3}+q+{\frac {k^{3}}{z^{3}}}=0}
z 3 + q − p 3 27 z 3 = 0 {\displaystyle z^{3}+q-{\frac {p^{3}}{27z^{3}}}=0}
z 6 + q z 3 − 27 p 3 = 0 {\displaystyle z^{6}+qz^{3}-27p^{3}=0}
由是，得 z 3 {\displaystyle z^{3}} 乃 y 2 + q y − p 3 27 = 0 {\displaystyle y^{2}+qy-{\frac {p^{3}}{27}}=0} 解之一。以一元二次方程公式可解之。其解同上卡贊諾之法，緣此與彼同一方程也。
復以單位根求 z {\displaystyle z} 之三可能值，以 z + k z = z − p 3 z {\displaystyle z+{\frac {k}{z}}=z-{\frac {p}{3z}}} 得 t {\displaystyle t} 。遂以 x = t − b 3 a {\displaystyle x=t-{\frac {b}{3a}}} 得解也。
先設判別式： Δ = q 2 4 + p 3 27 = ( − b 3 27 a 3 + b c 6 a 2 − d 2 a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 {\displaystyle \Delta ={\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}=(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}
則：
若 Δ > 0 {\displaystyle \Delta >0} ，有實根一且共軛複根一雙；
若 Δ = 0 {\displaystyle \Delta =0} ，有多重實根，其中：
若 Δ < 0 {\displaystyle \Delta <0} ，有三異實根。
![{\displaystyle x_{1}=-{\frac {b}{3a}}+{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\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}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7053397146c075f33c93932ee40050c90f465f25)
![{\displaystyle x_{2}=-{\frac {b}{3a}}+{\frac {-1-i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}+{\frac {-1+i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f8491659c0e85c95bc20abbab1b23945030226f)
![{\displaystyle x_{3}=-{\frac {b}{3a}}+{\frac {-1+i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}+{\frac {-1-i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1f5268f2fea7078ea40f6ba6759f822b4e513d7)
，以 
代之，得 
之制，其中：
之三可能值加諸 
成解也。 也。
，有實根一且共軛複根一雙；
，有多重實根，其中：
，有一二重實根且一單實根。
，有一三重實根。
，有三異實根。
