### Cubic Equations (Solution 1)

 Example for cube Express a'x3 + b'x2 + c'x + d' = 0 as y3 + cy + d = 0 Take the solution as u+v. Express (u+v)3 = f1(u,v)(u+v) + f2(u,v). (u+v)3 = 3uv(u+v) + u3+v3 solve f1 = 3uv = −c ----(i) f2 = u3 + v3 = −d ----(ii) Substituting value of u = −c/3v in (ii), we get (−c/3v)3 + v3 = −d −c3/27 + v6 = −dv3 v6 + dv3 − c3/27 = 0 solve the above equation considering it as a quadratic in v3 So v3 = −d/2 + √(d2/4 + c3/27) So, v = [−d/2 + √(d2/4 + c3/27)]1/3 then u = [−d/2 − √(d2/4 + c3/27)]1/3 The root of y3 + cy + d = 0 is y = u + v The root of the original equation (a'x3 + b'x2 + c'x + d' = 0) is −b'/3a' + u + v where c = (27a'2d' +2b'3 −9a'b'c')/(27a'3) and d = (c'/a' − b'2/3a'2) x = −b'/3a' + u + v = −b'/3a' + [−d/2 + √(d2/4 + c3/27)]1/3 + [−d/2 − √(d2/4 + c3/27)]1/3 a'x3 + b'x2 + c'x + d' = 0 is equivalent to y3 + [c'/a' − b'2 /3a'2]y + [(27a'2d' +2b'3 −9a'b'c')/(27a'3)] = 0 where y = x + b'/3a' or y3 + cy + d = 0 where c = (c'/a' − b'2/3a'2) and d = (27a'2d' +2b'3 −9a'b'c')/(27a'3) The cube root of u3 is u, ωu and ω2u. The cube root of v3 is v, ωv and ω2v. where 1, ω and ω2 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 + √(d2/4 + c3/27)]1/3 + [−d/2 − √(d2/4 + c3/27)]1/3    = −b'/3a' + ω[−d/2 + √(d2/4 + c3/27)]1/3 + ω2[−d/2 − √(d2/4 + c3/27)]1/3    = −b'/3a' + ω2[−d/2 + √(d2/4 + c3/27)]1/3 + ω[−d/2 − √(d2/4 + c3/27)]1/3 The roots can be written as x = −b'/3a' + u − c/3u