Master the Factorization Formula for x^200 and y^200 | Homework Equations

AI Thread Summary
The discussion revolves around the factorization of the expression x^200 - y^200, which is represented as (x-y)(x^199 + x^198y + ... + y^199). Participants explore whether this factorization has a specific name or identity associated with it. While some express difficulty in remembering such identities, others note that the formula is straightforward and easily verifiable. The conversation also touches on the utility of calculators for factoring complex expressions, with mentions of the TI-89 model. Ultimately, the consensus is that while it may not have a formal name, the factorization is a simple and useful mathematical identity.
ehrenfest
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Homework Statement



Is there like a formula with a name easy to remember of which the following is a specific instance:

x^200 -y^200 = (x-y)(x^199+x^198y+... + y^198*x + y^199)

?

Homework Equations





The Attempt at a Solution

 
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Not that I've ever heard...

(x+y)(x-y)(x^2+y^2)(x^4+y^4)(x^4+x^3y+x^{2}y^2+xy^3+y^4)(x^4-x^3y+x^2y^2-xy^3+y^4)(x^8-x^6y^2+x^4y^4-x^2y^6+y^8)(x^{16}-x^{12}y^4+x^8y^8-x^4y^{12}+y^{16})
(x^{20}+x^{15}y^5+x^{10}y^{10}+x^5y^{15}+y^{20})(x^{20}-x^{15}y^5+x^{10}y^{10}-x^5y^{15}+y^{20})(x^{40}-x^{30}y^{10}+x^{20}y^{20}-x^{10}y^{30}+y^{40})
(x^{80}-x^{60}y^{20}+x^{40}y^{40}-x^{20}y^{60}+y^{80})

Sorry wouldn't fint onto one line...

Maybe to can find a pattern to that mess though. If not, well, it took my calculator about 3 second to find the answer, might want to use one the next time you need to factor that...
 
Last edited:
which calculator do you have? i been thinking about upgrading to a TI-89 ...
 
rocophysics said:
which calculator do you have? i been thinking about upgrading to a TI-89 ...

Ti-89 Titanium
 
ehrenfest said:

Homework Statement



Is there like a formula with a name easy to remember of which the following is a specific instance:

x^200 -y^200 = (x-y)(x^199+x^198y+... + y^198*x + y^199)

?

Do you mean
x^n- y^n= (x-y)(x^{n-1}+ x^(n-2)y+ /cdot/cdot/cdot+ xy^{n-2}+ y^{n-1}?
I've never worried about it having a name!
 
HallsofIvy said:
Do you mean
x^n- y^n= (x-y)(x^{n-1}+ x^(n-2)y+ /cdot/cdot/cdot+ xy^{n-2}+ y^{n-1}?
I've never worried about it having a name!

Yes. I just forget identities so easily when I have to remember them by their statement not some label. However, you're right, this is a pretty simple one that is easily verifiable and probably doesn't warrant a name!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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