一元三次方程

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 ${\displaystyle ax^{3}+bx^{2}+cx+d=0}$。

立方函數 ${\displaystyle f(x)={\frac {x^{3}}{4}}+{\frac {3x^{2}}{4}}-{\frac {3x}{2}}-2}$ 嘅圖。佢嘅實根（即係 ${\displaystyle x}$ 截距）有三個： ${\displaystyle x=-4}$ ， ${\displaystyle x=-1}$ 及 ${\displaystyle x=2}$；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： ${\displaystyle x_{1}=-{\frac {b}{3a}}+{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}+{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}$

${\displaystyle x_{2}=-{\frac {b}{3a}}+{\frac {-1-i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}+{\frac {-1+i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}$

${\displaystyle x_{3}=-{\frac {b}{3a}}+{\frac {-1+i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}+{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}+{\frac {-1-i{\sqrt {3}}}{2}}{\sqrt[{3}]{-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}}-{\sqrt {(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}}}}}$

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 ${\displaystyle a}$ ，代入 ${\displaystyle x=t-{\frac {b}{3a}}}$ ，得到 ${\displaystyle t^{3}+pt+q=0}$ 咁嘅樣，當中： ${\displaystyle p={\frac {c}{3a}}-{\frac {b^{2}}{3a^{2}}}}$

${\displaystyle q={\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}}$

證明： ${\displaystyle ax^{3}+bx^{2}+cx+d=0}$

${\displaystyle x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x+{\frac {d}{a}}=0}$

${\displaystyle (t-{\frac {b}{3a}})^{3}+{\frac {b}{a}}(t-{\frac {b}{3a}})^{2}+{\frac {c}{a}}(t-{\frac {b}{3a}})+{\frac {d}{a}}=0}$

${\displaystyle t^{3}-{\frac {b}{a}}t^{2}+{\frac {b^{2}}{3a^{2}}}t-{\frac {b^{3}}{27a^{3}}}+{\frac {b}{a}}t^{2}-{\frac {2b^{2}}{3a^{2}}}t+{\frac {b^{3}}{9a^{3}}}+{\frac {c}{a}}t-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}=0}$

${\displaystyle t^{3}+({\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}})t+({\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}})=0}$

${\displaystyle t^{3}+pt+q=0}$

最後嗰步就將 ${\displaystyle t}$ 嗰三個可能值加落 ${\displaystyle -{\frac {b}{3a}}}$ 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 ${\displaystyle t}$ 。

卡贊諾之法

數學家卡贊諾話：假設 ${\displaystyle t=u+v}$。咁就可以得到：

${\displaystyle (u+v)^{3}+p(u+v)+q=0}$

${\displaystyle u^{3}+v^{3}+(3uv+p)(u+v)+q=0}$

同時，又設 ${\displaystyle 3uv+p=0}$。於是就得到： ${\displaystyle {\begin{cases}u^{3}+v^{3}=-q...(1)\\u^{3}v^{3}=-{\frac {p^{3}}{27}}...(2)\\\end{cases}}}$

${\displaystyle u^{3}}$ 同埋 ${\displaystyle v^{3}}$ 都係 ${\displaystyle y^{2}+qy-{\frac {p^{3}}{27}}=0}$ 嘅解。用一元二次方程嘅公式就可以解咗佢。於是得到：

${\displaystyle u^{3},v^{3}={\frac {-q\pm {\sqrt {q^{2}+{\frac {4}{27}}p^{3}}}}{2}}=-{\frac {q}{2}}\pm {\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}$

跟著用單位根去揾 ${\displaystyle u,v}$ 各自相應嘅三個可能值，用 ${\displaystyle u+v}$ 可以揾到 ${\displaystyle t}$ （兩個實數解嘅 ${\displaystyle u}$ 同 ${\displaystyle v}$ 相加揾到一個 ${\displaystyle t}$，至於複數解嘅 ${\displaystyle u}$ 同 ${\displaystyle v}$，佢哋嘅系數都係共軛嘅就相加）。於是，再用 ${\displaystyle x=t-{\frac {b}{3a}}}$ 就可以揾到答案。

韋達之法

數學家韋達話：假設 ${\displaystyle t=z+{\frac {k}{z}}}$。咁就可以得到：

${\displaystyle (z+{\frac {k}{z}})^{3}+p(z+{\frac {k}{z}})+q=0}$

${\displaystyle z^{3}+(3k+p)(z+{\frac {k}{z}})+{\frac {k^{3}}{z^{3}}}+q=0}$

又：設 ${\displaystyle 3k+p=0}$。則： ${\displaystyle z^{3}+q+{\frac {k^{3}}{z^{3}}}=0}$

${\displaystyle z^{3}+q-{\frac {p^{3}}{27z^{3}}}=0}$

${\displaystyle z^{6}+qz^{3}-27p^{3}=0}$

咁嘅話，就得到 ${\displaystyle z^{3}}$ 其實係 ${\displaystyle y^{2}+qy-{\frac {p^{3}}{27}}=0}$ 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 ${\displaystyle z}$ 嘅三個可能值，然後又用 ${\displaystyle z+{\frac {k}{z}}=z-{\frac {p}{3z}}}$ 得到 ${\displaystyle t}$ 。最後用 ${\displaystyle x=t-{\frac {b}{3a}}}$ 得到答案。

根嘅特性

首先定立判別式： ${\displaystyle \Delta ={\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}=(-{\frac {b^{3}}{27a^{3}}}+{\frac {bc}{6a^{2}}}-{\frac {d}{2a}})^{2}+({\frac {c}{3a}}-{\frac {b^{2}}{9a^{2}}})^{3}}$

咁樣嘅話：

如果 ${\displaystyle \Delta >0}$，就有一個實根同埋一啤共軛複根；

如果 ${\displaystyle \Delta =0}$，就有多重實根，當中：

• 如果 ${\displaystyle {\frac {q^{2}}{4}}={\frac {p^{3}}{27}}\neq 0}$，有一個雙重實根同埋另一個單實根。
• 如果 ${\displaystyle {\frac {q^{2}}{4}}={\frac {p^{3}}{27}}=0}$，有一個三重實根。

如果 ${\displaystyle \Delta <0}$，有三個唔同嘅實根。