Another
method of solving a cubic polynomial equation submitted independently by Paul
A. Torres and Robert A. Warren. It is based on the idea of "completing the
cube," by arranging matters so that three of the four terms are three of the
four terms of a perfect cube.
Start with
the cubic equation
x3 + ax2 + bx + c = 0.
If a2
- 3b = 0, then the first three terms are the first three terms of a perfect
cube, namely (x+a/3)3. Then you can "complete the cube" by
subtracting c from both sides and adding the missing term of the cube a3/27
to both sides. Recalling that b = a2/3, you get
x3 + ax2 + bx + c = 0,
x3 + ax2 + (a2/3)x + c = 0,
x3 + ax2 + (a2/3)x = -c,
x3 + ax2 + (a2/3)x + a3/27 =
a3/27 - c,
(x+a/3)3 = a3/27 - c.
By taking
the cube root of the left side and the three cube roots of the right side, you
get
x = -a/3 + (a3/27 - c)1/3,
x = -a/3 + (a3/27 - c)1/3(-1+sqrt[-3])/2,
x = -a/3 + (a3/27 - c)1/3(-1-sqrt[-3])/2.
These are
the roots of the cubic equation that were sought.
If a2
- 3b is nonzero, then proceed as follows. Set x = y + z, where y is an
indeterminate and z is a function of a, b, and c, which will be found below.
Then
(y+z)3 + a(y+z)2 + b(y+z) + c = 0,
y3 + (3z+a)y2 + (3z2+2az+b)y + (z3+az2+bz+c)
= 0,
y3 + dy2 + ey + f = 0,
where
d = 3z + a,
e = 3z2 + 2az + b,
f = z3 + az2 + bz + c.
The first
three terms of this equation in y will be those of a perfect cube if and only
if d2 - 3e is zero, which happens if and only if a2 - 3b
= 0, which cannot happen in this case, so we seemingly haven't gained anything.
However, the last three terms of this equation in y will be those of a
perfect cube if and only if e2 = 3df, that is if and only if
(3z2+2az+b)2 = 3(3z+a)(z3+az2+bz+c),
(a2-3b)z2 +
(ab-9c)z + (b2-3ac) = 0,
gz2 + hz + i = 0,
where
g = a2 - 3b,
h = ab - 9c,
i = b2 - 3ac.
Since a2
- 3b is nonzero, g is nonzero, and we have a true quadratic equation, called
the resolvent quadratic. Now we pick z to be a root of this quadratic equation.
If the GCD of z2 + (h/g)z + (i/g) and z3 + az2
+ bz + c as polynomials in z is not 1, then any root of the GCD is also a root
of the original cubic equation in x. Once you have at least one root, the
problem of finding the other roots is reduced to solving a quadratic or linear equation.
If the GCD is 1, then neither value of z can make f = 0, so we can assume
henceforth that f is nonzero. Either root z of the quadratic will do, but we
must choose one of them. We arbitrarily pick the one with a plus sign in front
of the radical,
z = (-h+sqrt[h2-4gi])/(2g),
= (9c-ab+sqrt[81c2-54abc+12a3c+12b3-3a2b2])/(2[a2-3b]).
Set z equal
to this value in the equation for y, and divide it by f on both sides. Then the
last three terms of the cubic in y are those of a perfect cube, namely (ey/[3f]+1)3,
so we can complete the cube to solve it. We do this by subtracting y3/f
from both sides, then adding the missing term of the cubic, (ey/[3f])3
to both sides, obtaining
(ey/[3f]+1)3 = ([e/(3f)]3-1/f)y3,
ey/(3f) + 1 = y([e/(3f)]3-1/f)1/3,
ey/(3f) + 1 = y([e/(3f)]3-1/f)1/3(-1+sqrt[-3])/2,
ey/(3f) + 1 = y([e/(3f)]3-1/f)1/3(-1-sqrt[-3])/2,
y = 3f/[-e+(e3-27f2)1/3],
y = 3f/[-e+(e3-27f2)1/3(-1+sqrt[-3])/2],
y = 3f/[-e+(e3-27f2)1/3(-1-sqrt[-3])/2].
Now you
have the values of y. Add z to each to get the values of x:
x = z + 3f/[(e3-27f2)1/3-e],
x = z + 3f/[(e3-27f2)1/3(-1+sqrt[-3])/2-e],
x = z + 3f/[(e3-27f2)1/3(-1-sqrt[-3])/2-e].
These are
the roots of the cubic equation that were sought.
Example: Solve x3 + 6x2 + 9x + 6 = 0.
a = 6, b = 9, c = 6.
Then
d = 3z + 6, e = 3z2 +
12z + 9, f = z3 + 6z2
+ 9z + 6.
Then
g = a2 - 3b = 36 - 27 = 9,
h = ab - 9c = 54 - 54 = 0,
i = b2 - 3ac = 81 - 108 = -27,
and the resolvent quadratic is
9z2 - 27 = 0,
z2 - 3 = 0,
z = sqrt(3).
Then
d = 6 + 3 sqrt(3),
e = 18 + 12 sqrt(3),
f = 24 + 12 sqrt(3),
and the cubic in y is
y3 + (6+3 sqrt[3])y2 + (18+12 sqrt[3])y + (24+12
sqrt[3]) = 0.
Then one root is
y = 3f/[(e3-27f2)1/3-e],
= 3(24+12 sqrt[3])/([18+12
sqrt(3)]3-27[24+12 sqrt(3)]2)1/3-18-12
sqrt[3]),
= (12+6 sqrt[3])/([9+6 sqrt(3)]1/3-3-2
sqrt[3]),
= (6+4 sqrt[3])/([2+sqrt(3)]1/3-2-sqrt[3]).
After a lot of simplification, you get
y = -2 - sqrt(3) - (2-sqrt[3])1/3 - (2+sqrt[3])1/3,
x = -2 - (2-sqrt[3])1/3 - (2+sqrt[3])1/3
(and two other roots).