What is the General Method for Deriving Cubic Roots?

[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 into 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

 

[AndyCav]

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

 

[AndyCav]

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

Ha ha, I appear to have found out!

 

[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

 

[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.

 

[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.

 

[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.