Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/254

25. Maintenant voici les valeurs de $\alpha$ pour les équations $x^{n}-1=0$ depuis $n=1$ jusqu’à $n=6\,:$

$n=2\left\{{\begin{aligned}\alpha \ =&-1,\\\alpha ^{2}=&1\,;\end{aligned}}\right.$

$n=3\left\{{\begin{aligned}\alpha \ =&{\frac {-1+{\sqrt {-3}}}{2}},\\\alpha ^{2}=&{\frac {-1-{\sqrt {-3}}}{2}},\\\alpha ^{3}=&1\,;\end{aligned}}\right.$

$n=4\left\{{\begin{aligned}\alpha \ =&{\sqrt {-1}},\\\alpha ^{2}=&-1,\\\alpha ^{3}=&-{\sqrt {-1}},\\\alpha ^{4}=&1\,;\end{aligned}}\right.$

$n=5\left\{{\begin{aligned}\alpha \ =&{\frac {{\sqrt {5}}-1}{4}}+{\frac {\sqrt {10+2{\sqrt {5}}}}{4}}{\sqrt {-1}},\\\alpha ^{2}=&-{\frac {{\sqrt {5}}+1}{4}}+{\frac {\sqrt {10-2{\sqrt {5}}}}{4}}{\sqrt {-1}},\\\alpha ^{3}=&-{\frac {{\sqrt {5}}+1}{4}}-{\frac {\sqrt {10-2{\sqrt {5}}}}{4}}{\sqrt {-1}},\\\alpha ^{4}=&{\frac {{\sqrt {5}}-1}{4}}-{\frac {\sqrt {10+2{\sqrt {5}}}}{4}}{\sqrt {-1}},\\\alpha ^{5}=&1\,;\end{aligned}}\right.$

$n=6\left\{{\begin{aligned}\alpha \ =&{\frac {1+{\sqrt {-3}}}{2}},\\\alpha ^{2}=&{\frac {-1+{\sqrt {-3}}}{2}},\\\alpha ^{3}=&-1,\\\alpha ^{4}=&{\frac {-1-{\sqrt {-3}}}{2}},\\\alpha ^{5}=&{\frac {1-{\sqrt {-3}}}{2}},\\\alpha ^{6}=&1.\end{aligned}}\right.$