Example for cubeExpress a'x ^{3} + b'x^{2} + c'x + d' = 0 as y^{3} + cy + d = 0Take the solution as u+v. Express (u+v) ^{3} = f_{1}(u,v)(u+v) + f_{2}(u,v).(u+v) ^{3} = 3uv(u+v) + u^{3}+v^{3}solve f _{1} = 3uv = −c ----(i)f _{2} = u^{3} + v^{3} = −d ----(ii)Substituting value of u = −c/3v in (ii), we get (−c/3v) ^{3} + v^{3} = −d −c ^{3}/27 + v^{6} = −dv^{3} v ^{6} + dv^{3} − c^{3}/27 = 0solve the above equation considering it as a quadratic in v ^{3}So v ^{3} = −d/2 + √(d^{2}/4 + c^{3}/27)So, v = [−d/2 + √(d ^{2}/4 + c^{3}/27)]^{1/3}then u = [−d/2 − √(d ^{2}/4 + c^{3}/27)]^{1/3}
The root of y
^{3} + cy + d = 0 is y = u + vThe root of the original equation (a'x ^{3} + b'x^{2} + c'x + d' = 0) is −b'/3a' + u + v where c = (27a' ^{2}d' +2b'^{3} −9a'b'c')/(27a'^{3}) andd = (c'/a' − b' ^{2}/3a'^{2})x = −b'/3a' + u + v = −b'/3a' + [−d/2 + √(d ^{2}/4 + c^{3}/27)]^{1/3} + [−d/2 − √(d^{2}/4 + c^{3}/27)]^{1/3}^{3} + b'x^{2} + c'x + d' = 0 is equivalent to y ^{3} + [c'/a' − b'^{2} /3a'^{2}]y + [(27a'^{2}d' +2b'^{3} −9a'b'c')/(27a'^{3})] = 0 wherey = x + b'/3a' or y ^{3} + cy + d = 0where c = (c'/a' − b' ^{2}/3a'^{2}) andd = (27a' ^{2}d' +2b'^{3} −9a'b'c')/(27a'^{3})
The cube root of u
^{3} is u, ωu and ω^{2}u.The cube root of v ^{3} is v, ωv and ω^{2}v.where 1, ω and ω ^{2} are cube roots of unity.As 3uv = −c i.e. the product of u and v is real So,the roots are x = −b'/3a' + u + v = −b'/3a' + [−d/2 + √(d ^{2}/4 + c^{3}/27)]^{1/3} + [−d/2 − √(d^{2}/4 + c^{3}/27)]^{1/3}= −b'/3a' + ω[−d/2 + √(d ^{2}/4 + c^{3}/27)]^{1/3} + ω^{2}[−d/2 − √(d^{2}/4 + c^{3}/27)]^{1/3}= −b'/3a' + ω ^{2}[−d/2 + √(d^{2}/4 + c^{3}/27)]^{1/3} + ω[−d/2 − √(d^{2}/4 + c^{3}/27)]^{1/3}The roots can be written as x = −b'/3a' + u − c/3u |

My discoveries > Polynomial Equations >