五十二角形
解と係数の関係より
u 3 − x 1 u 2 − 13 u − x 4 = 0 {\displaystyle u^{3}-x_{1}u^{2}-{\sqrt {13}}u-x_{4}=0}
変数変換
u = v + x 1 / 3 {\displaystyle u=v+x_{1}/3}
整理すると
v 3 − 13 + 3 13 6 v − ( 13 + 6 13 ) x 1 + 27 x 4 27 = 0 {\displaystyle v^{3}-{\frac {13+3{\sqrt {13}}}{6}}v-{\frac {(13+6{\sqrt {13}})x_{1}+27x_{4}}{27}}=0}
三角関数、逆三角関数を用いた解は
u 1 = x 1 3 + 2 3 13 + 3 13 2 cos  ( 1 3 arccos  ( ( 13 + 6 13 ) x 1 + 27 x 4 ( 13 + 3 13 ) 13 + 3 13 2 ) ) {\displaystyle u_{1}={\frac {x_{1}}{3}}+{\frac {2}{3}}{\sqrt {\frac {13+3{\sqrt {13}}}{2}}}\cos \left({\frac {1}{3}}\arccos \left({\frac {(13+6{\sqrt {13}})x_{1}+27x_{4}}{(13+3{\sqrt {13}}){\sqrt {\frac {13+3{\sqrt {13}}}{2}}}}}\right)\right)}