Solving Cubic Equations M W
A general cubic equation:
ax3 + bx2 + cx + d = 0 (a 6= 0) (1) reduces, after one divides both sides by a, to the equivalent equation
x3 + b0x2 + c0x + d0 = 0 () where b c d b0 ˜ , c0 ˜ , and d0 ˜ . () a a a 3 0 2 b0 Note that x + b x is equal to the cube of x + 3 up to a linear term in x:
b0 3 b0 2 b0 3 x3 + b0x2 = x + − 3 x − . () 3 3 3
Thus we obtain
b0 3 (b0)2 b0 3 x3 + b0x2 + c0x + d0 = x + + c0 − + d0 − 3 3 3 b0 3 (b0)2 b0 b0c0 2(b0)3 = x + + c0 − x + + d0 − + 3 3 3 3 27 = ξ 3 + c00ξ + d00 after substituting b0 ξ ˜ x + () 3 and setting (b0)2 b0c0 2(b0)3 c00 ˜ c0 − and d00 ˜ d0 − + . () 3 3 27 The reduction of the general cubic equation, (1), to the equation without a quadratic term: ξ 3 + c00ξ + d00 = 0 ()
1 was an easy part. In order to solve equation (), we represent ξ as the sum of two new unknowns ξ = u + v. () Note that
ξ 3 + c00ξ = u3 + v3 + 3uv(u + v) + c00(u + v) = u3 + v3 + (3uv + c00)ξ = u3 + v3 precisely when u and v satisfy equality
c00 uv = − . () 3 Then equation () becomes u3 + v3 + d00 = 0. (1) Let us multiply both sides of equation (1) by u3 and use equality (). What we get is the following particularly simple equation in u of degree :
c00 3 u6 − d00u3 − = 0 (11) 3 which is a quadratic equation in u3. Due to symmetry between u and v we can select u3 to be s d00 d00 2 c00 3 d00 √ u3 = − + + = − + ∆ (1) 2 2 3 2 and v3 to be s d00 d00 2 c00 3 d00 √ v3 = − − + = − − ∆ (1) 2 2 3 2 where we put d00 2 c00 3 ∆ ˜ + . (1) 2 3 Then, at least formally, r d00 √ u = 3 − + ∆, (1) 2
and r d00 √ v = 3 − − ∆, (1) 2 and finally r r d00 √ d00 √ x = 3 − + ∆ + 3 − − ∆. (1) 2 2 Now, one has to use the above formulae very cautiously, especially the last one. If we exclude the case c00 = 0 which renders equation () trivial:
ξ 3 + d00 = 0, (1) then any u and v satisfying condition () are nonzero. In particular, u3 and v3 are nonzero. A nonzero (complex) number, or more generally an element of any field in which any cubic equation has a solution, has exactly three different roots. Let u0 be any cubic root of d00 √ − + ∆. (1) 2 The remaining two roots are
2 −1 u1 = ζuo and u2 = ζ u0 = ζ u0 () where ζ is a nontrivial, i.e., not equal 1, cubic root of 1. There are two completely symmetric choices for ζ: √ √ −1 + −3 −1 − −3 and . (1) 2 2 They correspond to the two possible choices for the square root of −3. It does not depend which choice we make. If we choose one to be ζ then the other is ζ −1. Having discussed all three choices for the cubic root of (1), the corresponding three choices for the cubic root of d00 √ − − ∆ () 2 are 00 00 00 c c −1 c v0 = − , v1 = − = ζ v0, and v2 = − ζv0. () 3u0 3u1 3u2
The final result is this complete set of solutions of the original equation (1):
b0 c00 b0 x0 = u0 + v0 − = u0 − − , () 3 3u0 3 where u0 is an arbitrarily chosen cubic root of (1),
0 00 −1 0 −1 b c ζ b x1 = u0ζ + v0ζ − = u0ζ − − , () 3 3u0 3 and 0 00 0 − b −1 c ζ b x2 = u0ζ−1 + v0ζ = u0ζ − − . () 3 3u0 3 Note that the number of possible choices for u0 is , not , in general: two choices for the square root of ∆, assuming ∆ is nonzero, and three choices for the cubic root of (1) when the c00 is nonzero. Now, it is important to understand that any of these six choices will produce all three roots of equation (1). As we have seen, any particular choice assigns labels , 1, and to these roots, but when we move through all choices for u0, the labels , 1, and , are assigned to the roots differently each time! You are seeing here the Galois group of a general cubic equation in action! It would take us too far to attempt here an introduction to Galois Theory. Let me just say that the Galois group of a general cubic polynomial acts as the group of all permutations of the set of roots {x0,x1,x2}, and is therefore icomorphic to the symmetric group S3 that you have encountered in your Introduction to Abstract Algebra.