Completing the cube!!!

anonimnystefy · 2012-01-19 10:16:30

hi bobbym

here's the thread you asked for.can we continue here?

bobbym · 2012-01-19 10:25:11

We sure can as long as you change the title to the correct one.

anonimnystefy · 2012-01-19 10:28:15

could you change it.i don't know how i wrote square instead of cube.

btw,is there a way for me to change it?

bobbym · 2012-01-19 10:29:40

You could try editing the first post. But I have already done it for you.

anonimnystefy · 2012-01-19 10:32:01

cannot be done that way.

anyway,...back to the matter at hand.where are those text files?

bobbym · 2012-01-19 10:37:45

Hi anonimnystefy;

I have copied the text file you have requested. Inconsistent bracketing and a missed bracket make this suspect. I have tried to clean it up but I could only guess where the missing bracket should go.

Another method of solving a cubic polynomial equation submitted independently by Paul A. Torres and Robert A. Warren. It is based on the idea of "completing the cube," by arranging matters so that three of the four terms are three of the four terms of a perfect cube.
Start with the cubic equation

If

then the first three terms are the first three terms of a perfect cube, namely

Then you can "complete the cube" by subtracting c from both sides and adding the missing term of the cube

to both sides. Recalling that

you get:

By taking the cube root of the left side and the three cube roots of the right side, you get:

These are the roots of the cubic equation that were sought.

If

then proceed as follows. Set x = y + z, where y is an indeterminate and z is a function of a, b, and c, which will be found below. Then:

where

The first three terms of this equation in y will be those of a perfect cube iff

which happens iff

which cannot happen in this case, so we seemingly haven't gained anything. However, the last three terms of this equation in y will be those of a perfect cube iff

that is iff

where

Since

then

and we have a true quadratic equation, called the resolvent quadratic. Now we pick z to be a root of this quadratic equation.

If

then any root of the GCD is also a root of the original cubic equation in x. Once you have at least one root, the problem of finding the other roots is reduced to solving a quadratic or linear equation.

If

then neither value of z can make f = 0, so we can assume henceforth that f is nonzero. Either root z of the quadratic will do, but we must choose one of them. We arbitrarily pick the one with a plus sign in front of the radical:

Set z equal to this value in the equation for y, and divide it by f on both sides. Then the last three terms of the cubic in y are those of a perfect cube, namely:

so we can complete the cube to solve it. We do this by subtracting

from both sides, then adding the missing term of the cubic,

to both sides, obtaining

Now you have the values of y. Add z to each to get the values of x:

These are the roots of the cubic equation that were sought.

Example:

We have a = 6, b = 9, c = 6.

Then

The resolvent quadratic is

the cubic in y is

Then one root is

After a lot of simplification, you get

And two other roots that he does not provide. I checked the one he has given and it is correct.

Math Is Fun Forum

#1 2012-01-19 10:16:30

Completing the cube!!!

#2 2012-01-19 10:25:11

Re: Completing the cube!!!

#3 2012-01-19 10:28:15

Re: Completing the cube!!!

#4 2012-01-19 10:29:40

Re: Completing the cube!!!

#5 2012-01-19 10:32:01

Re: Completing the cube!!!

#6 2012-01-19 10:37:45

Re: Completing the cube!!!

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