Deriving general cube roots

by AndyCav
Tags: cube, deriving, roots
AndyCav is offline
Apr21-05, 10:57 AM
P: 7
I've been trying to derive general solutions for cubic roots, i.e. the general solutions of (will Latex just work?)


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!

Phys.Org News Partner Mathematics news on
Math modeling handbook now available
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
AndyCav is offline
Apr21-05, 11:00 AM
P: 7
PS Anyone know how some peole are making Latex code work in their messages?
AndyCav is offline
Apr21-05, 11:04 AM
P: 7

Ha ha, I appear to have found out!

AndyCav is offline
Apr21-05, 11:18 AM
P: 7

Deriving general cube roots

This is what I tried:

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

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

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

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

Now, as [tex]x_0[/tex] is by definition a solution of f; [tex]f(x_0)=0[/tex] we get:

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


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

AndyCav is offline
Apr21-05, 11:26 AM
P: 7
There's meant to be a sqaure root sign over that -c/a but it's not coming out on my screen.

[tex]x_0[/tex] is of course just one solution of the cubic.
CRGreathouse is offline
Apr21-05, 11:34 AM
Sci Advisor
HW Helper
P: 3,680
If you have one solution [tex]x_0[/tex] to the cubic, you can just divide by [tex]x-x_0[/tex] 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 [tex]bx^2[/tex] term by a change of variable.
AndyCav is offline
Apr21-05, 12:28 PM
P: 7
Well I'm trying to derive all solutions generally so I don't have [tex]x_0[/tex] 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.

Register to reply

Related Discussions
Modular cube roots Linear & Abstract Algebra 14
calculating cube roots General Math 16
cube roots Precalculus Mathematics Homework 13
'Euler criterion' for cube roots? Linear & Abstract Algebra 1
cube roots Calculus & Beyond Homework 11