Cubic Equations - Another Solution

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### Another Solution of the Cubic Equation

Here is 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.

```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,
```
```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).