解方程

northwolves

实数范围内解方程 $x^3+1=2\root3(2x-1)$

显然$x_{1}=1$

wayne

$-16 x + 3 x^3 + 3 x^6 + x^9 + 9=(-1+x) (-1+x+x^2) (9+2 x+4 x^2+2 x^3+2 x^4+x^6)=0$
而$9+2 x+4 x^2+2 x^3+2 x^4+x^6>0$恒成立、

northwolves

wayne 发表于 2021-1-26 22:56
$-16 x + 3 x^3 + 3 x^6 + x^9 + 9=(-1+x) (-1+x+x^2) (9+2 x+4 x^2+2 x^3+2 x^4+x^6)=0$
而$9+2 x+4 x^2 ...

只能硬算？有没有简单巧妙的方法？

wayne

灵感乍现，我怎么感觉这题目我做过。
$x^3+1=2y, y^3+1=2x$， 而$y=1/2(x^3-1)$是单调递增函数， 因为轮换对称，所以该函数关于$y=x$对称，反函数等于函数本身，所以 $x^3+1=2x$，也就是$(-1+x) (-1+x+x^2)=0$

mathematica

1. (*解方程https://bbs.emath.ac.cn/thread-17672-1-1.html*)
   
2. Clear["Global`*"];(*mathematica11.2,win7(64bit)Clear all variables*)
   
3. ans=Solve[(x^3+1)^3==8*(2*x-1),{x}];
   
4. Grid[ans,Alignment->Left]
   
5. Grid[N@ans,Alignment->Left]
   

复制代码

\[\begin{array}{l}
x\to 1 \\
x\to \frac{1}{2} \left(-\sqrt{5}-1\right) \\
x\to \frac{1}{2} \left(\sqrt{5}-1\right) \\
x\to \text{Root}\left[\text{$\#$1}^6+2 \text{$\#$1}^4+2 \text{$\#$1}^3+4 \text{$\#$1}^2+2 \text{$\#$1}+9\&,1\right] \\
x\to \text{Root}\left[\text{$\#$1}^6+2 \text{$\#$1}^4+2 \text{$\#$1}^3+4 \text{$\#$1}^2+2 \text{$\#$1}+9\&,2\right] \\
x\to \text{Root}\left[\text{$\#$1}^6+2 \text{$\#$1}^4+2 \text{$\#$1}^3+4 \text{$\#$1}^2+2 \text{$\#$1}+9\&,3\right] \\
x\to \text{Root}\left[\text{$\#$1}^6+2 \text{$\#$1}^4+2 \text{$\#$1}^3+4 \text{$\#$1}^2+2 \text{$\#$1}+9\&,4\right] \\
x\to \text{Root}\left[\text{$\#$1}^6+2 \text{$\#$1}^4+2 \text{$\#$1}^3+4 \text{$\#$1}^2+2 \text{$\#$1}+9\&,5\right] \\
x\to \text{Root}\left[\text{$\#$1}^6+2 \text{$\#$1}^4+2 \text{$\#$1}^3+4 \text{$\#$1}^2+2 \text{$\#$1}+9\&,6\right] \\
\end{array}\]

数值化
\[\begin{array}{l}
x\to 1. \\
x\to -1.61803 \\
x\to 0.618034 \\
x\to -1.08891-0.824866 i \\
x\to -1.08891+0.824866 i \\
x\to 0.123133\, -1.4302 i \\
x\to 0.123133\, +1.4302 i \\
x\to 0.965776\, -1.18647 i \\
x\to 0.965776\, +1.18647 i \\
\end{array}\]

mathematica

老问题：
https://bbs.emath.ac.cn/forum.ph ... 290&fromuid=865