一元三次方程式

解法

1. $a t^{3} + b t^{2} + c t + d = 0 \Rightarrow$ 以 $t = x - \frac{b}{3 a}$ 代入

2.得 $x^{3} + p x + q = 0$ ，令其三根為 $x_{1}, x_{2}, x_{3}$

3.

令 ${\begin{matrix} x_{1} = u + v \\ x_{2} = u ω + v ω^{2} \\ x_{3} = u ω^{2} + v ω \end{matrix}$	$\Rightarrow$	$(x - x_{1}) (x - x_{2}) (x - x_{3}) = 0$	$\Rightarrow$	${\begin{matrix} x_{1} + x_{2} + x_{3} = 0 \\ x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = - 3 u v = p \\ - x_{1} x_{2} x_{3} = - (u^{3} + v^{3}) = q \end{matrix}$

4.

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

5.故 $y_{1}, y_{2}$ 為 $y^{2} + q y + - (\frac{p}{3})^{3} = 0$ 的兩根

6.

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

題目： $x^{3} - 3 x + 2 = 0$

題目： $x^{3} - 12 x - 16 = 0$