Cubing the Cubic Formula Help would be greatly appreciated

In summary, the conversation discussed the cubic formula and its relationship to the quadratic formula. The speaker shared their discovery of being able to rearrange the quadratic formula to solve it in a different way, and attempted to do the same with the cubic formula. They also mentioned the use of Galois theory to prove the unsolvability of quintics and higher degree polynomials.
  • #1
Frogeyedpeas
80
0
Hi,

So I was working on a little project the other day and it was in regards to the cubic formula...

Which can be found here:

http://en.wikipedia.org/wiki/Cubic_formula#General_formula_of_roots

basically given an equation of the form:

ax3 + bx2 + cx + d = 0 the formulas on the attached link will give you the three values of x that satisfy this equation.

Here is my dilemma,

I was just kind of playing around with the quadratic equation which is:

x = [itex]\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/itex]

It can be conveniently "re arranged" to look like this:

(2ax + b)2 = b2 - 4ac

So in dealing with the cubic equation I tried the same thing which was fairly simple:

(3ax + b) = Ω (where Ω is all of our other business that remains)...

Can somebody tell me the simplified value of Ω3?

Its terribly difficult to solve by hand and I do not own a symbolic calculator can manipulate it so if somebody could do the solution that would be convenient...

Additionally I would like to know another value:

Ω can be seen as the sum of (σ + μ) (see the two giant cube roots in the wiki article, those are sig and mu for this expression).

What is (σ2 - σμ + μ2)?

and What is (σ3 - μ3)

Hopefully between those three questions I can resolve to find a pattern between the quadratic and cubic formulas.

Thanks!
 
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  • #2
I won't go through the detail of something that is in all the books just some overall ideas.

Frogeyedpeas said:
I was just kind of playing around with the quadratic equation which is:

x = [itex]\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/itex]

It can be conveniently "re arranged" to look like this:

(2ax + b)2 = b2 - 4ac

Playing around is good. You have discovered a little thing you had probably forgotten (often happens). That actually if quadratic equation solving was explained to you, it goes in the reverse order - you don't get the second formula from the first, you get the first from the second, which is the quadratic to be solved rearranged. You try to incorporate its first two terms into a perfect square and then you have a constant, (b2 - 4ac) left over. So you can solve taking square roots.

Put another way you get the difference of two squares

(x + b/2)2 - {√[(b2 - 4ac)/4a2]}2 = 0

which, taking a term to the other side and taking square roots leads to two linear equations.

For the cubic you try to do the same idea (to answer your question) - you try to express it as the difference of two cubes, e.g. if we make the leading term 1, then try to express any cubic with this leading term as

(x + β)3 - [(1 - α3)x3 + γ)3.

When you expand and try to identify coefficients with those of a cubic you find you need to solve a quadratic equation.

If the cubic has three real roots the roots of this quadratic are nonreal. It was the genius idea you could use them anyway and through them get the real roots of the cubic that was the takeoff that made math something else.

You can do the quartic too as difference of two squares, one being the square of a square; that involves solving a cubic equation.

Higher degree than that you can rearrange polynomials in pretty ways but this does into allow you to solve them, it is well known
 
  • #3
what is the proof that quintics and higher are unsolvabe? (i'm not familiar with galois theory or lie groups so I'm going to need it slightly more explained)
 
  • #4
Frogeyedpeas said:
what is the proof that quintics and higher are unsolvabe? (i'm not familiar with galois theory or lie groups so I'm going to need it slightly more explained)

Unfortunately, I don't think there is a proof that doesn't involve Galois theory.
 
  • #5
could u give me the proof (with theory) and i'll see if i can grasp something about it
 
  • #6
the best explanation i know of the cubic formula is in euler's elements of algebra. it is free on the web.
 

FAQ: Cubing the Cubic Formula Help would be greatly appreciated

What is the Cubic Formula and how does it work?

The Cubic Formula is a mathematical equation used to solve cubic equations, or equations with a variable raised to the third power. It works by using the coefficients of the equation to find the roots, or solutions, of the equation.

Why is cubing the cubic formula important in mathematics?

Cubing the cubic formula is important because it allows us to solve cubic equations, which are commonly found in many areas of mathematics, science, and engineering. It also serves as a foundation for understanding more complex mathematical concepts.

Can you provide an example of how to use the Cubic Formula?

Yes, an example of using the Cubic Formula to solve the equation x^3 + 2x^2 - 5x + 6 = 0 would be:x = -1 + (1 + 81)^(1/3) + (1 - 81)^(1/3)x = -1 + 4 + (-4)x = -1 + 4 - 4x = -1

Are there any limitations to using the Cubic Formula?

Yes, there are some limitations to using the Cubic Formula. It can only be used to solve cubic equations, not higher or lower degree equations. Additionally, the solutions may involve complex numbers, which can be difficult to interpret.

Is there a simpler method for solving cubic equations?

Yes, there are other methods for solving cubic equations, such as factoring or using the Rational Root Theorem. These methods may be more efficient and easier to understand for certain equations. However, the Cubic Formula is a reliable method for finding all possible solutions to a cubic equation.

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