# Deriving general cube roots

1. Apr 21, 2005

### AndyCav

I've been trying to derive general solutions for cubic roots, i.e. the general solutions of (will Latex just work?)

$ax^3+bx^2+cx+d=0$

I do not want to be shown the solutions - but does anyone know what direction to go in to achieve this?

I thought I'd found solutions at one point but they relied on one of the solutions (the simplest in form) being $\sqrt{-c/a}$ and that doesn't appear to be true!

Andy

2. Apr 21, 2005

### AndyCav

PS Anyone know how some peole are making Latex code work in their messages?

3. Apr 21, 2005

### AndyCav

$$x_{0}=\sqrt{\frac{-c}{a}}$$

Ha ha, I appear to have found out!

4. Apr 21, 2005

### AndyCav

This is what I tried:

$$f(x)=ax^3+bx^2+cx+d=0$$ is the general equation to solve, and intersects the y axis a distance d from the origin.

So, consider $$f(x-x_0)=g(x)$$ where $$x_0$$ is one solution of $$f(x)=0$$. As we have shifted the original function along the x-axis the new function $$g(x)$$ now passes through the origin.

After substituting $$x-x_0$$ into f to get g, expand g to get it in the form $$g(x)=px^3+qx^2+rx+s$$. As said above g passes through the origin so that s=0.

Therefore, $$s=0=d-ax_0^3+bx_0^2-cx_0$$. Rearranging for d in terms of $$x_0$$ and a, b and c and substituting back into the original form for f we get $$f(x)=a(x^3-x_0^3)+b(x^2-x_0^2)+c(x-x_0)$$.

Now, as $$x_0$$ is by definition a solution of f; $$f(x_0)=0$$ we get:

$$ax_0^3+cx_0=0$$ which gives the trivial solution $$x_0=0$$ (i.e. d=0) and the two solutions

$$x_0=+-\sqrt{\frac{-c}{a}}$$.

These don't seem to work. Am I doing something stupid?

Andy

Last edited: Apr 21, 2005
5. Apr 21, 2005

### AndyCav

There's meant to be a sqaure root sign over that -c/a but it's not coming out on my screen.

$$x_0$$ is of course just one solution of the cubic.

6. Apr 21, 2005

### CRGreathouse

If you have one solution $$x_0$$ to the cubic, you can just divide by $$x-x_0$$ to reduce it to a quadratic and you're done.

If you really want a general solution like Cardanno's, I think the first step would be to remove the $$bx^2$$ term by a change of variable.

7. Apr 21, 2005

### AndyCav

Well I'm trying to derive all solutions generally so I don't have $$x_0$$ to begin with. That's a fantastic idea about the change of variable to remove the even term though....! Cheers! I'll have dinner then have another try.

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