How Do You Prove the Difference of Powers Formula?

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Homework Help Overview

The discussion revolves around proving the difference of powers formula, specifically the expression x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1}). The original poster attempts to understand the proof process and the validity of their approach through distribution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods for proving the formula, including polynomial long division and induction. The original poster questions whether their distribution method constitutes a valid proof and seeks clarification on the proof process.

Discussion Status

Some participants provide guidance on using polynomial long division and suggest that the original poster's approach is acceptable. Multiple methods are being explored, including the possibility of treating the right-hand side as a geometric series, although there is some concern about circular reasoning in that approach.

Contextual Notes

There is mention of potential assumptions regarding the proof methods that can be used, as well as a request for clarification on formatting issues in LaTeX within the discussion.

Magicalz
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Homework Statement


Prove the following:

[tex]x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})[/tex]

The Attempt at a Solution


Ugh I just tried to distribute the right part:

[tex] \begin{equation*}<br /> \begin{split}<br /> \x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1}) \\<br /> &= \(x - y)x^{n-1} + (x - y)x^{n-2}y + ... + (x - y)xy^{n-2} + (x - y)y^{n-1} \\<br /> &= x^n - x^{n-1}y + x^{n-1}y - x^{n-2}y^2 + ... + x^2y^{n-2} - xy^{n-1} + xy^n-1 - y^n <br /> \end{split}<br /> \end{equation*}[/tex]

well the terms that are visible (that are written down) really do cancel out (except of course x^n and y^n) however I guess what I'm having trouble grasping with is what constitutes the "proof" part of this...I mean yes it seems likely that the terms inbetween x^n and y^n are cancelling each other out, but can we be absolutely sure? the dots ... mean continue the pattern,but I guess what I'm wondering is,is what I'm doing by distributing the right side constituting the "proof"? Any help is appreciated thankyou.
 
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Do you know anything about polynomial long division ? If so, regard the x^n -y^n and x-y as polynomials in the "x" variable. EDIT: [/tex]
 
Hi dextercioby! Thanks for the quick reply. Hmm I don't remember polynomial long division can you show the steps for this particular question? thanks!
 
Yes, you can prove it the way you did. Show that the RHS equals the LHS by performing the multiplications in the RHS. Totally okay. Or you can do it the way i suggested.

[tex]\frac{x^{n}-y^{n}}{x-y}=x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}[/tex]

By polynomial long division you can show that the LHS equals the RHS.
 
Hey dextercioby thanks for the help, can you give me a short reminder of how to do polynomial long division??

*oh and just another question, how come on my above attempted solution part, the line is so long, but i put the coding \\ in the latex but there seems to be no new line thing? can anyone help me thankyou
 
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You could also try induction, though its quite unnecessary.
 
[tex]\frac{x^{n}-y^{n}}{x-y}=x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}[/tex]

Treat the RHS as a geometric series, first term [tex]x^{n-1}[/tex], common ratio [tex]\frac{y}{x}[/tex]. Sum. Simplify.
 
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Kind of circular unless you know of a proof for the formula of the sum of a geometric series that doesn't exploit this identity, which i don't >.<
 
  • #10
Gib Z said:
Kind of circular unless you know of a proof for the formula of the sum of a geometric series that doesn't exploit this identity, which i don't >.<

I suppose you're right - but the question never stated what you could or could not assume. In any case, in addition to your suggestion of induction, it's just another approach one could use.

The proof for GP sum that I know (in fact I found it myself as a kid) is the obvious Sn = ..., rSn = ..., take the difference, etc.

One could also adapt that method to this, although it takes a little more intuition to see that one needs to multiply by (y/x) to get the result.
 

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