====== 三次方程式 ======
$$x^3+ax^2+bx+c=0 $$
===== 標準化 =====
$$(x-r)^3+a(x-r)^2+b(x-r)+c=0$$
$$x^3-3rx^2+3r^2x-r^3+ax^2-2arx+ar^2+bx-br+c=0$$
$$x^3+(a-3r)x^2+(3r^2-2ar+b)x+ar^2-r^3-br+c=0$$
===== 1実根 =====
$$ x^3+3px-2q=0 $$
$$ x=(q+\sqrt{D})^{\frac{1}{3}}+(q-\sqrt{D})^{\frac{1}{3}} $$

\begin{align}
x^3&=(q+\sqrt{D})+3(q+\sqrt{D})^{\frac{2}{3}}(q-\sqrt{D})^{\frac{1}{3}}+3(q+\sqrt{D})^{\frac{1}{3}}(q-\sqrt{D})^{\frac{2}{3}}+(q-\sqrt{D})\\
&=2q+3(q+\sqrt{D})^{\frac{1}{3}}(q-\sqrt{D})^{\frac{1}{3}}\left((q+\sqrt{D})^{\frac{1}{3}}+(q-\sqrt{D})^{\frac{1}{3}}\right)\\
&=2q+3(q^2-D)^{\frac{1}{3}}\left((q+\sqrt{D})^{\frac{1}{3}}+(q-\sqrt{D})^{\frac{1}{3}}\right)\\
\end{align}

$$(q^2-D)^{\frac{1}{3}}=-p$$
$$ D=q^2+p^3 $$

===== 因数分解 =====
 
$$x^3+3px-2q=(x-\alpha)(x-\beta)(x-\gamma)$$

\begin{align}
\alpha+\beta+\gamma=0\\
\alpha\beta+\beta\gamma+\gamma\alpha=3p\\
\alpha\beta\gamma=2q\\
\end{align}

$$\alpha+\beta=-\gamma$$
$$\alpha\beta=3p-(\alpha+\beta)\gamma=3p+\gamma^2$$
$$x^3+3px-2q=(x^2+\gamma x+3p+\gamma^2)(x-\gamma)$$