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


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