# 一元三次方程

## 解法

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

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

### 簡化

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

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

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

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

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

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

### 卡贊諾之法

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

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

${\displaystyle u^{3}}$ 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：

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

### 韋達之法

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

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

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

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

## 根嘅特性

• 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。