#"A general solution method for cubic equations can help here."#
#"I used a method based on the substitution of Vieta."#
#"Dividing by the first coefficient yields :"#
#v^3 + (500000/127) v^2 + (194000000/127) v + (1300000000/127) = 0#
#"Substituting v=y+p in "v^3+a v^2+b v+c" yields :"#
#y^3 + (3p+a) y^2 + (3p^2+2ap+b) y + p^3+ap^2+bp+c = 0#
#"if we take "3p+a=0" or "p=-a/3", the"#
#"first coefficients becomes zero, and we get :"#
#y^3 - (176086000000/48387) y + (139695127900000000/55306341) = 0#
#"(with "p = -500000/381")"#
#"Substituting "y=qz" in "y^3 + b y + c = 0", yields :"#
#z^3 + b z / q^2 + c / q^3 = 0#
#"if we take "q = sqrt(|b|/3)", the coefficient of z becomes 3 or -3,"#
#"and we get :"#
#"(here "q = 1101.38064036")"#
#z^3 - 3 z + 1.89057547 = 0#
#"Substituting "z = t + 1/t", yields :"#
#t^3 + 1/t^3 + 1.89057547 = 0#
#"Substituting "u = t^3", yields the quadratic equation :"#
#u^2 + 1.89057547 u + 1 = 0#
#"The roots of the quadratic equation are complex."#
#"This means that there are 3 real roots in our cubic equation"#
#"and that we need to use De Moivre's formula to take the"#
#"cube root in the solving process, which complicates matters."#
#"A root of this quadr. eq. is "u=-0.94528773 + 0.3262378 i.#
#"Substituting the variables back, yields :"#
#t = root3(u) = 1.0*(cos(-0.93642393)+i sin(-0.93642393))#
#= 0.59267214 - 0.80544382 i.#
#=> z = 1.18534427.#
#=> y = 1305.51523196.#
#=> x = -6.82072605.#
#"The other roots can be found by dividing and solving the"# #"remaining quadratic equation."#
#"They are : "-3501.59623563" and "-428.59091234.#
