代數方程 x 7 = 1 {\displaystyle x^{7}=1} 存在李煌根式解
x 7 = 1 {\displaystyle x_{7}=1}
x 1 = 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 − 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 {\displaystyle x_{1}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}}}
x 2 = 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 − 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 {\displaystyle x_{2}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}}}
x 3 = 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{3}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
x 4 = 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{4}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
x 5 = 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{5}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
x 6 = 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{6}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
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<<School:李煌數學研究院
存在李煌根式解
