一元三次方程，或者叫立方函數，係一條有一個未知數而次數係三嘅方程。一元三次方程通常會寫做 a x 3 + b x 2 + c x + d = 0 {\displaystyle ax^{3}+bx^{2}+cx+d=0} 。 立方函數 f ( x ) = x 3 4 + 3 x 2 4 − 3 x 2 − 2 {\displaystyle f(x)={\frac {x^{3}}{4}}+{\frac {3x^{2}}{4}}-{\frac {3x}{2}}-2} 嘅圖。佢嘅實根（即係 x {\displaystyle x} 截距）有三個： x = − 4 {\displaystyle x=-4} ， x = − 1 {\displaystyle x=-1} 及 x = 2 {\displaystyle x=2} ；佢嘅臨界點有兩個。 解法 如果要解一元三次方程，首先一定要剷咗二次方嗰個項，變成一個冇咗二次項嘅方程。之後無論用卡贊諾或者韋達嘅方法，都可以得到方程嘅解。最後得到嘅解嘅公式係： 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}}}}}} 簡化 如果想解一元三次方程，一定要先剷走咗個二次項。呢個係求解嘅時候好重要嘅一步，咁就可以啟發思路，容易啲諗到求解嘅辦法。簡化嘅時候，成條式除以 a {\displaystyle a} ，代入 x = t − b 3 a {\displaystyle x=t-{\frac {b}{3a}}} ，得到 t 3 + p t + q = 0 {\displaystyle t^{3}+pt+q=0} 咁嘅樣，當中： 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}}} 證明： 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} 最後嗰步就將 t {\displaystyle t} 嗰三個可能值加落 − b 3 a {\displaystyle -{\frac {b}{3a}}} 嗰度，咁就好以得到方程嘅三個解。 去到呢一步之後，無論用卡贊諾嘅方法或者韋達嘅方法，都可以揾到 t {\displaystyle t} 。 卡贊諾之法 數學家卡贊諾話：假設 t = u + v {\displaystyle t=u+v} 。咁就可以得到： ( 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} 同時，又設 3 u v + p = 0 {\displaystyle 3uv+p=0} 。於是就得到： { 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}} 同埋 v 3 {\displaystyle v^{3}} 都係 y 2 + q y − p 3 27 = 0 {\displaystyle y^{2}+qy-{\frac {p^{3}}{27}}=0} 嘅解。用一元二次方程嘅公式就可以解咗佢。於是得到： 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}}}}} 跟著用單位根去揾 u , v {\displaystyle u,v} 各自相應嘅三個可能值，用 u + v {\displaystyle u+v} 可以揾到 t {\displaystyle t} （兩個實數解嘅 u {\displaystyle u} 同 v {\displaystyle v} 相加揾到一個 t {\displaystyle t} ，至於複數解嘅 u {\displaystyle u} 同 v {\displaystyle v} ，佢哋嘅系數都係共軛嘅就相加）。於是，再用 x = t − b 3 a {\displaystyle x=t-{\frac {b}{3a}}} 就可以揾到答案。 韋達之法 數學家韋達話：假設 t = z + k z {\displaystyle t=z+{\frac {k}{z}}} 。咁就可以得到： ( 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} 咁嘅話，就得到 z 3 {\displaystyle z^{3}} 其實係 y 2 + q y − p 3 27 = 0 {\displaystyle y^{2}+qy-{\frac {p^{3}}{27}}=0} 嘅其中一個解。用一元二次方程嘅公式就可以揾到解。揾佢嘅解就好似上面卡贊諾嘅辦法，因為上面嗰條方程同埋下面嗰條方程都係一樣嘅。 之後再用單位根揾 z {\displaystyle z} 嘅三個可能值，然後又用 z + k z = z − p 3 z {\displaystyle z+{\frac {k}{z}}=z-{\frac {p}{3z}}} 得到 t {\displaystyle t} 。最後用 x = t − b 3 a {\displaystyle x=t-{\frac {b}{3a}}} 得到答案。 根嘅特性 首先定立判別式： 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}} 咁樣嘅話： 如果 Δ > 0 {\displaystyle \Delta >0} ，就有一個實根同埋一啤共軛複根； 如果 Δ = 0 {\displaystyle \Delta =0} ，就有多重實根，當中： 如果 q 2 4 = p 3 27 ≠ 0 {\displaystyle {\frac {q^{2}}{4}}={\frac {p^{3}}{27}}\neq 0} ，有一個雙重實根同埋另一個單實根。 如果 q 2 4 = p 3 27 = 0 {\displaystyle {\frac {q^{2}}{4}}={\frac {p^{3}}{27}}=0} ，有一個三重實根。 如果 Δ < 0 {\displaystyle \Delta <0} ，有三個唔同嘅實根。 Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.
![{\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}}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/7053397146c075f33c93932ee40050c90f465f25)
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，代入 
，得到 
咁嘅樣，當中：
嗰三個可能值加落 
嗰度，咁就好以得到方程嘅三個解。 。
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