十三角形
2 cos  4 π 13 + 2 cos  6 π 13 = − 2 + − 260 − 156 3 i 3 + − 260 + 156 3 i 3 6 = 1 3 ( − 1 + 13 ⋅ − 5 − 3 3 i 2 13 3 + 13 ⋅ − 5 + 3 3 i 2 13 3 ) {\displaystyle 2\cos {\frac {4\pi }{13}}+2\cos {\frac {6\pi }{13}}={\frac {-2+{\sqrt[{3}]{-260-156{\sqrt {3}}i}}+{\sqrt[{3}]{-260+156{\sqrt {3}}i}}}{6}}={\frac {1}{3}}\left(-1+{\sqrt {13}}\cdot {\sqrt[{3}]{\frac {-5-3{\sqrt {3}}i}{2{\sqrt {13}}}}}+{\sqrt {13}}\cdot {\sqrt[{3}]{\frac {-5+3{\sqrt {3}}i}{2{\sqrt {13}}}}}\right)}