# Solving cubics

1. Nov 19, 2005

### hypermonkey2

Hiya. Is anyone familiar with solving cubics? Currently, i am able to to solve cubics where
$$q^2 - p^3>0$$
however, i cannot when it is negative or equal to 0. what is the procedure?
ps. the p and q are from cardanos method. im assuming anyone who is familiar with the method knows what they represent.

2. Nov 19, 2005

### robert Ihnot

You have to recognize the fact of THE IRREDUCIBLE CUBIC. There is a lot of literature on that.
Now to get off that immediate subject but to show something about cubics, well, to make a short story long, this guy Niederhoffer, author of "Education of a Speculator," wanted to make his high school math team and was asked to evaluate, where p =1/3:
$$(2+\sqrt{5})^p+(2-\sqrt{5})^p$$

Neiderhoffer said he didn't have the time to figure it out, but guess right anyway.

Last edited: Nov 19, 2005
3. Nov 19, 2005

### hypermonkey2

interesting. that can be done without guessing using de moivres theorem perhaps no? in any case, how does this help me? hehe.

4. Nov 21, 2005

### hypermonkey2

Alright, so what are general methods that some of you use to solve cubic polynomials?

5. Nov 21, 2005

### amcavoy

I'd like to know this too. The only one I've seen is on Wolfram's MathWorld where you make many substitutions and try to kill the x2 term. It doesn't seem particularly difficult, but very time-consuming.

I asked a professor about this once and he said that 3rd, 4th, and fifth (or higher) degree polynomials are studied in abstract algebra (Galois Theory I believe, although I don't know anything about it).

6. Nov 21, 2005

### redkimchi

IIRC, there exists a general formula or a procedure giving all roots of any cubic equation.
You can find it in any abstract algebra textbook.

7. Nov 21, 2005

### robert Ihnot

Well, with the cubic, there is the well known "Cardan method." Ax^3+Bx^2+Cx +D and eliminate the B term by the substitution: x=y-B/3A, leaving the "depressed cubic": y^3+py+q = 0. Then we substitute: y=z-p/3z and arrive at: Z^3-(p^3)/(27Z^3)+q, which allows us to solve for a quadratic in Z^3.

You can find most of this in places like "the Penguine Dictionary of Mathmatics" under "Cubic."

They are not so happy to call this the Cardan or "Cardano's method" these days since Cardan readily admitted to gaining the method from Tartaglia after promising to keep it a secret.

Interesting facts about Cardan: He admitted in his autobiography to having a real love for gambling. (See, Ore, "Cardano the Gambling Scholar." He was very brilliant and at one point considered the greatest physician in the world. He wrote many popular books and contributed to study of Algebra. He cast the horoscope of Jesus Christ, wrote a book in praise of Nero, and successfully predicted his own death. (Though his critics claim he committed suicide.)http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Cardan.html

Last edited: Nov 21, 2005
8. Nov 21, 2005

### hypermonkey2

Fascinating! This is exactly the method i have been studying. It works wonderfully! However, in some cases when we solve the resulting quadratic, we arrive with a negative discriminant. However, in this context, it can still result in Real roots for the cubic. It is this delicacy that perplexes me.

9. Nov 21, 2005

### hypermorphism

You already know the algebra of complex numbers ?

10. Nov 21, 2005

### robert Ihnot

Well, Galois gives the problem X^3-3X+1. Let omega = the cubic root of 1:
$$\omega=\frac{1+i\sqrt3}{2}$$
Then Galois gives us $$x=\omega^(1/3) +\omega^(-1/3)$$
The first part can be written as:
(cos120+isin120)^1/3=cos40 + isin40. And the second part is just the conjugate cos40-isin40, so we get:
X=2cos40.
Now if somebody were to ask, "WHY does this work?", well avoiding all complexity, one could say, "You can not trisect the angle."
Thus we have no real number expression in surds for cos40, so to get the answer we need to use De Moivre's Theorem.

PS: For the sake of completeness, if we are looking for the other roots, well when we have the form:
A^1/3 + B^1/3, by the cubic roots of unity we have the three cases: 1, w, w^2, where w =omega, the cube root of one.

So looking at distinct cases (one part multiplied by w, the other by w^2) we have also: w^4/3+w^5/3, which reduces to 2cos160=-2cos20. (In this case cos160=cos200=-cos20.) And 2cos80.

So we have three real solutions: -2cos20, 2cos40, 2cos80.

Last edited: Nov 22, 2005
11. Nov 21, 2005

### mathwonk

For the solutions x1, x2 of the quadratic equation x^2+px+q = 0, we have (since at least 825 AD) the formula x = (1/2){-p + D^(1/2)}, in terms of the coefficients p, q, where D = (x1-x2)^2 = p^2-4q. The two solutions are obtained by taking the two square roots of D. For the cubic equation x^3+px+q = 0, years of toil and some intrigue led to the publication, by Cardano in 1545, of the following formula. x =

(1/3)[{(-27q/2)+(3/2)(-3D)^(1/2)}^(1/3) - 3p/{(-27q/2) + (3/2)(-3D)^(1/2)}^(1/3)].

Fixing a value of (-3D)^(1/2), where
D = (x1- x2)^2(x1-x3)^2(x2-x3)^2 = -4p^3-27q^2, and varying the cube root gives all three solutions. Eg., in the equation x^3-1 = 0, p = 0, q = -1, D = -27, so we get x = (1/3){27/2 + 27/2}^(1/3) = {1/2 + 1/2}^(1/3) = 1^(1/3) = 1, as hoped.

Similarly, for x^3-a = 0, we have p = 0, q = -a, D = -27a^2, hence
x = (1/3){27a/2 + (3/2)(81a^2)^(1/2)}^(1/3) = (1/3){27a}^(1/3) = a^(1/3).

For x^3 - 4x = 0, we get p = -4, q = 0, D = 256, and so

x = (1/3) [{(3/2)(-768)^(1/2)}(1/3) + 12/{(3/2)(-768)^(1/2)}^(1/3)]

= (1/3) [ {(-27)(64)}^(1/6) + 12/{(-27)(64)}^(1/6) ]

= (1/3) [ 2sqrt(3) i^(1/3) + 12/{2sqrt(3) i^(1/3)} ] = (2/sqrt(3) )( i^(1/3) + i^(-1/3))

= (4/sqrt(3) )Re(i^(1/3)).

Varying the cube roots of i in this formula gives
(4/sqrt(3) )(cos(