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== 李煌公式 ==
* 方程''x''<sup>''n''</sup>=1滿足''n'' =3k+2, ''n''>3,n,k是整數,之李煌根式解爲：
*: <math>x_{1}={\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}+{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}} } </math>
*: <math>x_{2}={\Bigg(\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}+{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}}\Bigg )}^{2}</math>
*: <math>x_{3}={\Bigg(\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}+{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}}\Bigg )}^{3}</math>
*: <math>...</math>
*: <math>x_{n}={\Bigg(\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}+{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}}\Bigg )}^{n}=1</math>

* 方程''x''<sup>''n''</sup>=1滿足''n'' =3k+1, ''n''>3,n,k是整數,之李煌根式解爲：
*: <math>x_{1}={\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}-{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}} } </math>
*: <math>x_{2}={\Bigg(\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}-{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}}\Bigg )}^{2}</math>
*: <math>x_{3}={\Bigg(\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}-{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}}\Bigg )}^{3}</math>
*: <math>...</math>
*: <math>x_{n}={\Bigg(\frac{{-\frac{1}{2}}-{\frac{i\sqrt{3}}{2}}  }{(-\frac{1}{2}-{\frac{i\sqrt{3}}{2})}^{\frac{1}{n}}}\Bigg )}^{n}=1</math>

== 來源 ==
* 《南昌理工學院學報》.李煌

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