# Finding roots to cubics

1. Oct 16, 2005

### uob_student

hello

how can i finding roots to cubics??

explain by example

2. Oct 16, 2005

### Tide

Perhaps you have an example in mind?

3. Oct 16, 2005

### mezarashi

Oh how I just love these open-ended questions ;P

4. Oct 16, 2005

### Fermat

If you want to find approximate vaues, use newton's method of approximation. (you can google for it).
For a more analytical approach try here

5. Oct 16, 2005

### uob_student

x^3+(5*x^2)+(3*x)+9=0

6. Oct 16, 2005

### uob_student

no i do not want newton's method of approximation

i want (kardan) method but i am not sure about the spelling of (kardan)

7. Oct 16, 2005

### Tide

Cardan's formula gives

$$x = -\frac {\left( 179 - 9 \sqrt {345} \right)^{1/3}}{3}-\frac {\left( 179 + 9 \sqrt {345} \right)^{1/3}}{3} - \frac {5}{3}$$

for the real root. The other two are complex. And, no, I did not do it by hand! :)

8. Oct 16, 2005

### HallsofIvy

Staff Emeritus
If a and b are any two numbers then
(a- b)3= a3-3a2b+ 3ab2- b3
3ab(a-b)= 3a2b- 3ab2

so (a-b)3+ 3ab(a-b)= a3- b3.

In particular, if we let x= a-b, m= 3ab, and n= a3- b3, that says that x3+ mx= n. That is, we can pick any two numbers a, b and right down a cubic equation that has x= a- b as a root.
The question is, can we go the other way: given m and n, can we find a and b so we can write x= a-b as a solution.
The answer to that question is "Yes, we can"!

Since m= 3ab, b= m/3a. Putting that int n= a3- b3, we have $$n= a^3- \frac{m^3}{3^3a^3}$$.
Multiplying both sides of the equation by a3, we have
$$na^3= a^6- (\frac{m}{3})^3$$
which looks worse but is just a quadratic equation in a3:
$$(a^3)^2- n(a^3)- (\frac{m}{3})^3$$.
Use the quadratic formula to solve that
$$a^3= \frac{n +/- \sqrt{n^2- 4\left(\frac{m}{3}\right)^3}}{2}$$
$$a^3= \frac{n}{2} +/- \sqrt{\left(\frac{n}{2} \right)^2- \left(\frac{m}{3}\right)^3}$$

Since a3- b3= n, solving for b3 gives
$$a^3= -\frac{n}{2} +/- \sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}$$

Finding the cube root of each of those, then subtracting to get x= a- b gives the formula that Tide cited.

Warning- applying that formula is really, really hard!

Last edited by a moderator: Oct 16, 2005
9. Oct 16, 2005

thanks