# A math question

1. Jul 19, 2008

### glueball8

Is there a easy way to solve x^3-x-1=0? I know there's newton's method and depressed method... but is there a easy way?

2. Jul 19, 2008

### mathman

Third degree and fourth degree polynomial equations have exact solutions in terms of radicals. With a little searching, you can find them.

3. Jul 19, 2008

### glueball8

Hmm its not that simple.... you can't use long division or the |_____.....

its a special polynomial equation..

its from trying to solve (1+(1+(1+...)^1/3)^1/3)^1/3...

4. Jul 19, 2008

### n_bourbaki

It is that simple: there are solutions for cubics in terms of the coefficients and sums and radicals thereof. Just google for them.

5. Jul 19, 2008

### glueball8

6. Jul 20, 2008

### gel

7. Jul 20, 2008

### glueball8

can you explain how?

8. Jul 20, 2008

### gel

ok.
Actually, because this cubic has only one real root, you should use the hyperbolic cosine instead.
The triple angle formula is $\cosh(3\theta)=4\cosh^3\theta-3\cosh\theta$. Try putting $x=a\cosh\theta$ into your formula for x.
$$a^3\cosh^3(\theta)-a\cosh(\theta)=1.$$
notice if a2 = 4/3 then you can multiply both sides by 3/a,
$$4\cosh^3(\theta)-3\cosh(\theta)=3/a=3\sqrt{3}/2$$
Putting this together, you have $x = \frac{2}{\sqrt{3}}\cosh\left(\cosh^{-1}(3\sqrt{3}/2)/3\right)$.

9. Jul 20, 2008

### Daniel Y.

Hmm? What's so hard about it? The constant term is -1, so chances are x - 1 or x +1 divides the polynomial. A little synthetic division magic and we see x-1 divides the cubic without remainder. Since x - 1 is a factor, we know:

$$(x^3 -x -1) = (x-1)(x^2 +x+1) = 0$$

So $$x-1 = 0$$ and or $$x^2 +x +1 = 0$$ (I assume you know how to solve quadratic equations)

Little method I use for solving polynomials with an order > 2 is:

1) Figure out which factors there might be of the equation. Usually in x^n +- x^(n-1)...+-C there is a factor x - c such that c is a factor of C. For instance, in $$x^4 - x^3 + x^2 - 8$$ possible factors could be x +- 2, +-4, or +-8.

2) Divide the polynomial by the possible factors until you find one without a remainder.

3) Repeat step 2 until you 'take out' all the factors.

4) Try the answers, because some roots don't always work!

10. Jul 21, 2008

### HallsofIvy

Staff Emeritus
No.

Apparently this is harder than you think it is because x= 1 is obviously NOT a root. 13- 1- 1= -1, not 0.
(x-1)(x2+ x+ 1)= x3+ x2+ x- x2- x- 1= x3- 1, not x3- x- 1.

11. Jul 22, 2008

### uart

Yes, unless you can just happen to guess one real solution (as per Daniel Y's attempt, it's always worth attempting this first --- although it appears Daniel made an arithmetic error or something in this case and got it wrong) then that "depressed cubic" method is really the best way to go.

Personlly I like to do via a non-linear substitution. Fundamentally it's really the same thing as using those formulas you linked but I find it somehow more satisfying to use the substitution.

Given a depressed cubic $z^3 + pz + q = 0$, if you make the non-linear substition $z = w - p/(3w)$ then it "magically" simplifies into a quadratic in $w^3$. Give it a try, it's easy enough. Although be prepared to work in the complex domain for all the intermediate work, even for cases when all the final roots are real!

12. Jul 22, 2008

### glueball8

ya my teacher showed me that one... I thought it was too complex... and I didn't understand it...