两个无理数的和为1

Kingofforest

考虑如下方程:
  \[{x^3}+x-2 = 0\]
\[(x^2+x+2)(x-1) = 0\]

则该方程有实根\(x = 1\)

又令\[x = u + v\]
\[x^3+x-2 = (u+v)(3uv+1)+u^3+v^3-2 = 0\]
\(\begin{cases}
u^3v^3\ =\ -\frac{1}{27}\\
u^3+v^3=2
\end{cases}\)
若\(u^3\),\(v^3\)分别对应\(t_1\),\(t_2\)
则有\[t^2-2t-\frac{1}{27}=0\]
\[t = 1\pm\sqrt{\frac{28}{27}}\]
问是否有\[x_0 = u_0 + v_0 = \sqrt[3]{1+\sqrt{\frac{28}{27}}}+\sqrt[3]{1-\sqrt{\frac{28}{27}}}=1\]

hbghlyj

仿照1#方法由$(x-1)(x^2+ax+b)=0$解得 $$1=\frac{2^{\frac{1}{3}} {\left(2 \, a^{2} + 2^{\frac{1}{3}} {\left(-2 \, a^{3} - 3 \, a^{2} + 9 \, a b + 3 \, a + 18 \, b + \sqrt{-4 \, {\left(a^{2} + a - 3 \, b + 1\right)}^{3} + {\left(2 \, a^{3} + 3 \, a^{2} - 9 \, a b - 3 \, a - 18 \, b - 2\right)}^{2}} + 2\right)}^{\frac{2}{3}} + 2 \, a - 6 \, b + 2\right)}}{6 \, {\left(-2 \, a^{3} - 3 \, a^{2} + 9 \, a b + 3 \, a + 18 \, b + \sqrt{-4 \, {\left(a^{2} + a - 3 \, b + 1\right)}^{3} + {\left(2 \, a^{3} + 3 \, a^{2} - 9 \, a b - 3 \, a - 18 \, b - 2\right)}^{2}} + 2\right)}^{\frac{1}{3}}} - \frac{1}{3} \, a + \frac{1}{3}$$用SageMath对满足$b>a^2/4$的随机实数$a,b$验证上式：

输出1.00000000000000

Kingofforest

hbghlyj 发表于 2023-7-26 05:36
本帖最后由 hbghlyj 于 2023-7-26 05:51 编辑 仿照1#方法由$(x-1)(x^2+ax+b)=0$解得
$$1=\frac{2^{\frac{1 ...

太强了！

isee

这就是卡当公式，必然相等.

isee

isee 发表于 2023-7-27 21:47
这就是卡当公式，必然相等.

胆子再大一点，设 \[\sqrt[3]{1+\sqrt{\frac{28}{27}}}=\sqrt[3]{1+\frac29\sqrt{21}}=\frac12+m\cdot \sqrt{21},\]
两边立方，整理得到
\[1+\frac29\sqrt{21}=\frac{63m^2}2+\frac18+\Big(\frac{3m}4+21m^3\Big)\sqrt{21},\]
则需\[\frac{63m^2}2+\frac18=1,\Rightarrow m=\frac16,\]

下面只需检验当$m=1/6$时，$\frac{3m}4+21m^3$值为$2/9$即可.

亦即 \[\sqrt[3]{1\pm\sqrt{\frac{28}{27}}}=\frac12\pm\frac16\sqrt{21}.\]