有点难，求解韦神的一元三次方程

陈九章

mathe

设$x=y+1/y$代入

nyy

韦神个屁

1. Clear["Global`*"];(*Clear all variables*)
   
2. ans=Solve[x^3-3*x+1==0,{x}]//FullSimplify
   
3. Grid[ans,Alignment->Right](*列表显示*)
   

复制代码

\[\begin{array}{r}
x\to 2 \cos \left(\frac{2 \pi }{9}\right) \\
x\to \text{Root}\left[\text{$\#$1}^3-3 \text{$\#$1}+1\&,2\right] \\
x\to \text{Root}\left[\text{$\#$1}^3-3 \text{$\#$1}+1\&,1\right] \\
\end{array}\]

有一个根解出来了，剩下的简单了

\[\begin{array}{r}
x\to \sqrt[3]{\frac{1}{2} \left(-1+i \sqrt{3}\right)}+\frac{1}{\sqrt[3]{\frac{1}{2} \left(-1+i \sqrt{3}\right)}} \\
x\to -\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(-1+i \sqrt{3}\right)}-\frac{1-i \sqrt{3}}{2^{2/3} \sqrt[3]{-1+i \sqrt{3}}} \\
x\to -\frac{1}{2} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(-1+i \sqrt{3}\right)}-\frac{1+i \sqrt{3}}{2^{2/3} \sqrt[3]{-1+i \sqrt{3}}} \\
\end{array}\]

{{x -> -1.8793852415718167681}, {x -> 0.34729635533386069770}, {x -> 1.5320888862379560704}}

还不如数值解爽！

Jack315

方程为 \(x^3+px+q=0\)，其中：\(p=-3, q=1\)
第一个（实）根为：
\(x_1=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}\)
\(=\sqrt[3]{-\frac{1}{2}+\sqrt{(\frac{1}{2})^2+(\frac{-3}{3})^3}}+\sqrt[3]{-\frac{1}{2}-\sqrt{(\frac{1}{2})^2+(\frac{-3}{3})^3}}\)
\(=\sqrt[3]{-\frac{1}{2}+i\frac{\sqrt{3}}{2}}+\sqrt[3]{-\frac{1}{2}-i\frac{\sqrt{3}}{2}}\)
\(=(e^{i\frac{2}{3}\pi})^\frac{1}{3}+(e^{-i\frac{2}{3}\pi})^\frac{1}{3}\)
\(=2\cos(\frac{2}{9}\pi)\)
剩下两个根为下列一元二次方程的根：
\(x^2+x_1x+p+x_1^2=0\)
\(x_{2,3}=\frac{1}{2}(-x_1\pm\sqrt{x_1^2-4(p+x_1^2)})\)
\(=\frac{1}{2}(-2\cos(\frac{2}{9}\pi)\pm\sqrt{4\cos^2(\frac{2}{9}\pi)-4(-3+4\cos^2(\frac{2}{9}\pi))})\)
\(=-\cos(\frac{2}{9}\pi)\pm\sqrt{\cos^2(\frac{2}{9}\pi)-(-3+4\cos^2(\frac{2}{9}\pi))}\)
\(=-\cos(\frac{2}{9}\pi)\pm\sqrt{3}\sin(\frac{2}{9}\pi)\)

Jack315

【答案】
方程 \(x^3-3x+1=0\) 的三个根为：
\(\begin{Bmatrix}
2\cos(\frac{2}{9}\pi), & -2\sin(\frac{7}{18}\pi), & 2\sin(\frac{1}{18}\pi)
\end{Bmatrix}\)

Jack315

答案再写得优雅点是这样的：
\(\begin{matrix} x=2\sin(k\frac{\pi}{18}) & k=1,13,29 \end{matrix}\)

陈九章

谢谢mathe老师、nyy老师、Jack315老师！
代韦神向您们致敬! 韦神的解答如下：