What is the General Method for Deriving Cubic Roots?

In summary, the conversation discusses trying to find general solutions for cubic roots, specifically the general solutions of $ax^3+bx^2+cx+d=0$, without being shown the solutions. The conversation also mentions using Latex code to display equations and suggests a possible solution involving substituting x-x_0 into the original equation and using a change of variable to remove the even term.
  • #1
AndyCav
7
0
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
 
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  • #2
PS Anyone know how some peole are making Latex code work in their messages?
 
  • #3
[tex]x_{0}=\sqrt{\frac{-c}{a}}[/tex]

Ha ha, I appear to have found out!
 
  • #4
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

[tex]x_0=+-\sqrt{\frac{-c}{a}}[/tex].


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

Andy
 
Last edited:
  • #5
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.
 
  • #6
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.
 
  • #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.
 

What is the process of deriving general cube roots?

Deriving general cube roots involves finding the value of a number that, when cubed, equals the given number. This can be done through a process of trial and error or by using algebraic methods such as the cube root formula.

What is the cube root formula?

The cube root formula is a mathematical equation used to find the cube root of a given number. It is written as:
∛x = y, where y is the cube root of x.

How do you find the cube root of a perfect cube?

To find the cube root of a perfect cube, you can use the cube root formula or simply take the number and divide it by 3, since the cube root of a perfect cube is always an integer.

What is the difference between a real and complex cube root?

A real cube root is a number that, when cubed, equals a real number. A complex cube root is a number that, when cubed, equals a complex number. Complex cube roots have both a real and imaginary component, while real cube roots are purely real numbers.

Can you explain the concept of rationalizing the denominator in cube root expressions?

Rationalizing the denominator in cube root expressions involves eliminating any cube root from the denominator of a fraction. This is done by multiplying both the numerator and denominator by a cube root expression that will result in the cube root being eliminated from the denominator.

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