任何一元三次方程都可化简为 ax^3+bx^2+cx+d=0(a,b,c,d \in \mathbb R, 且 a \ne 0) 的形式。当 d=0 时，ax^3+bx^2+cx=0(a,b,c \in \mathbb R, 且 a \ne 0) 可以容易地进行因式分解降次。所以此文仅推导 ax^3+bx^2+cx+d=0(a,b,c,d \in \mathbb R, 且 a,d \ne 0) 的解法。
为了方便，我们设 w=\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i，则 w^{3k+1}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,w^{3k+2}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i,w^{3k}=1,k \in \mathbb Z。
\Large{b,c=0}
推导过程：
\begin{aligned}
ax^3+bx^2+c+d&=0\\
ax^3&=-d\\
x^3&=-\frac{d}{a}\\
x&=w^k\sqrt[3]{-\frac{d}{a}},k \in \{0,1,2\}
\end{aligned}得出公式：
\begin{aligned}
x_1&=\sqrt[3]{-\frac{d}{a}}\\
x_2&=w\sqrt[3]{-\frac{d}{a}}\\
x_3&=w^2\sqrt[3]{-\frac{d}{a}}
\end{aligned}\Large{b=0,c \ne 0}
推导过程：
\begin{aligned}
ax^3+bx^2+cx+d&=0\\
x^3+\frac{c}{a}x+\frac{d}{a}&=0
\end{aligned}令 x=y-\frac{c}{3ay}，得
\begin{aligned}
(y-\frac{c}{3ay})^3+\frac{c}{a}(y-\frac{c}{3ay})+\frac{d}{a}&=0\\
y^3-\frac{c}{a}y+\frac{c^2}{3a^2y}-(\frac{c}{3ay})^3+\frac{c}{a}y-\frac{c^2}{3a^2y}+\frac{d}{a}&=0\\
y^3+\frac{d}{a}-(\frac{c}{3a})^3\frac{1}{y^3}&=0\\
y^6+\frac{d}{a}y^3-(\frac{c}{3a})^3&=0
\end{aligned}由一元二次方程求根公式得
\begin{aligned}
y^3&=-\frac{d}{a}\div2\pm\sqrt{(\frac{d}{a})^2+4(\frac{c}{3a})^3}\div2\\
y^3&=-\frac{d}{2a}\pm\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}\\
y&=w^k\sqrt[3]{-\frac{d}{2a}+(-1)^k\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}},k \in \{0,1,2,3,4,5\}
\end{aligned}我们发现
\begin{aligned}
-\frac{c}{3a\sqrt[3]{-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}}&=-\frac{\frac{c}{3a}\sqrt[3]{(-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})^2}}{-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}\\
&=-\frac{\frac{c}{3a}\sqrt[3]{(-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})^2}(-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})}{(-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})(-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})}\\
&=-\frac{\frac{c}{3a}\sqrt[3]{(-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})^2(-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})^3}}{(\frac{d}{2a})^2-((\frac{d}{2a})^2+(\frac{c}{3a})^3)}\\
&=-\frac{\frac{c}{3a}\sqrt[3]{(-(\frac{c}{3a})^3)^2(-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3})}}{-(\frac{c}{3a})^3}\\
&=\frac{\sqrt[3]{(\frac{c}{3a})^6}\sqrt[3]{-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}}{(\frac{c}{3a})^2}\\
&=\frac{(\frac{c}{3a})^2\sqrt[3]{-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}}{(\frac{c}{3a})^2}\\
&=\sqrt[3]{-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}
\end{aligned}所以 x 仅有三个根，即
x=w^k\sqrt[3]{-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}+w^{-k}\sqrt[3]{-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}},k \in \{0,1,2\}得出公式：
\begin{aligned}
x_1&=\sqrt[3]{-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}+\sqrt[3]{-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}\\
x_2&=w\sqrt[3]{-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}+w^2\sqrt[3]{-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}\\
x_3&=w^2\sqrt[3]{-\frac{d}{2a}+\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}+w\sqrt[3]{-\frac{d}{2a}-\sqrt{(\frac{d}{2a})^2+(\frac{c}{3a})^3}}
\end{aligned}著名的卡丹公式：
方程 x^3+px+q=0 的解为
\begin{aligned}
x_1&=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}\\
x_2&=w\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}+w^2\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}\\
x_3&=w^2\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}+w\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}
\end{aligned}\Large{b \ne 0}
推导过程：
\begin{aligned}
ax^3+bx^2+cx+d&=0\\
x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}&=0
\end{aligned}令 x=y-\frac{b}{3a}，得
\begin{aligned}
(y-\frac{b}{3a})^3+\frac{b}{a}(y-\frac{b}{3a})^2+\frac{c}{a}(y-\frac{b}{3a})+\frac{d}{a}&=0\\
y^3-\frac{b}{a}y^2+\frac{b^2}{3a^2}y-\frac{b^3}{27a^3}+\frac{b}{a}y^2-\frac{2b^2}{3a^2}y+\frac{b^3}{9a^3}+\frac{c}{a}y-\frac{bc}{3a^2}+\frac{d}{a}&=0\\
y^3+\frac{3ac-b^2}{3a^2}y+\frac{2b^3+27a^2d-9abc}{27a^3}&=0
\end{aligned}由卡丹公式得
\begin{aligned}
y&=w^k\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}+\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}\\
&+w^{-k}\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}-\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}},k \in \{0,1,2\}
\end{aligned}于是
\begin{aligned}
x&=w^k\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}+\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}\\
&+w^{-k}\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}-\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}-\frac{b}{3a},k \in \{0,1,2\}
\end{aligned}得出公式：
\begin{aligned}
x_1&=\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}+\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}\\
&+\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}-\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}-\frac{b}{3a}\\
x_2&=w\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}+\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}\\
&+w^2\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}-\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}-\frac{b}{3a}\\
x_3&=w^2\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}+\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}\\
&+w\sqrt[3]{-\frac{2b^3+27a^2d-9abc}{54a^3}-\sqrt{(\frac{2b^3+27a^2d-9abc}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}-\frac{b}{3a}
\end{aligned}