# 一元三次方程式

## 解法

### 阶段一：变形去除三次项

1.${\displaystyle at^{3}+bt^{2}+ct+d=0\Rightarrow }$${\displaystyle t=x-{\frac {b}{3a}}}$ 代入

2.得 ${\displaystyle x^{3}+px+q=0}$，令其三根为 ${\displaystyle x_{1},x_{2},x_{3}}$

### 阶段二：变身为二次方程式

3.

${\displaystyle {\begin{cases}x_{1}=u+v\\x_{2}=u\omega +v\omega ^{2}\\x_{3}=u\omega ^{2}+v\omega \end{cases}}}$${\displaystyle \Rightarrow }$${\displaystyle (x-x_{1})(x-x_{2})(x-x_{3})=0}$${\displaystyle \Rightarrow }$${\displaystyle {\begin{cases}x_{1}+x_{2}+x_{3}=0\\x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}=-3uv=p\\-x_{1}x_{2}x_{3}=-(u^{3}+v^{3})=q\end{cases}}}$

4.

${\displaystyle {\begin{cases}y_{1}=u^{3}\\y_{2}=v^{3}\end{cases}}}$${\displaystyle \Rightarrow }$${\displaystyle {\begin{cases}y_{1}+y_{2}=-q\\y_{1}y_{2}=-({\frac {p}{3}})^{3}\end{cases}}}$

5.故 ${\displaystyle y_{1},y_{2}}$${\displaystyle y^{2}+qy+-({\frac {p}{3}})^{3}=0}$ 的两根

### 阶段三：以二次方程式之两根求三次方程式之三根

6.

${\displaystyle u={\sqrt[{3}]{y_{1}}},v={\sqrt[{3}]{y_{2}}}}$${\displaystyle {\begin{cases}x_{1}=u+v\\x_{2}=u\omega +v\omega ^{2}\\x_{3}=u\omega ^{2}+v\omega \end{cases}}}$

## 例题

### 例题一

1. ${\displaystyle y_{1},y_{2}}$${\displaystyle y^{2}+2y+1=0}$ 的两根，分别为 -1,-1，各开三方后分别为 -1,-1
2. ${\displaystyle x^{3}-3x+2=0}$ 的三根为 -1+-1,-1ω+-1ω2,-1ω2+-1ω，即 -2,1,1

### 例题二

1. ${\displaystyle y_{1},y_{2}}$${\displaystyle y^{2}-16y+64=0}$ 的两根，分别为 8,8，各开三方后分别为 2,2
2. ${\displaystyle x^{3}-3x+2=0}$ 的三根为 2+2,2ω+2ω2,2ω2+2ω，即 4,-2,-2