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I have questions about a formula that is :

$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$

That's how it's written on my textbook but it seems like I've trouble understanding it and so I guess it's why I'm struggling to show that is true...

My first question is what does the $\ldots$ represent really? I mean $n \geq 1$, so I don't understand for example if $n = 1$ do you "stop at $x^{n-1}$" so whenever you see a $0$ power or you stop at $x^n$?

Another question i have, is that even tho I see the "symmetry", like one term is suppressing another. Whenever i develop the right side, i of course have $x^n - y^n$ left but i get many other things like $x^2y^{n-2}$,$y^2x^{n-2}$, ect. I can't find a way that they could simplify themselves so where did I do wrong?

Thanks in advance and have a good day.

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  • $\begingroup$ For the case of $n=1$ you have $x-y = (x-y)(1)$. If this is unsettling to you, then formalize it as $x^n-y^n = (x-y)(\sum\limits_{k=0}^{n-1}x^{n-1-k}y^k)$ where there should be no confusion as to the contents of the second set of parentheses. $\endgroup$
    – JMoravitz
    Sep 23, 2021 at 14:27
  • $\begingroup$ See How acceptable is an ellipses "..." in formal mathematics?. $\endgroup$
    – JMoravitz
    Sep 23, 2021 at 14:34
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    $\begingroup$ As for proving that the assertion is accurate, explore the products $(x - y) \times (x + y)$ and $(x - y) \times (x^2 + xy + y^2)$ and $(x - y) \times (x^3 + x^2y + xy^2 + y^3).$ Then, after seeing the pattern re terms cancelling out, explore the general product $\displaystyle (x - y)\left[\sum_{k=0}^{n-1} x^{n-1-k}y^k\right].$ $\endgroup$
    – user2661923
    Sep 23, 2021 at 14:35
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    $\begingroup$ Does this answer your question? Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$ $\endgroup$
    – JMoravitz
    Sep 23, 2021 at 14:36
  • $\begingroup$ Yeah, it does and it's way more clear with a sum like that thanks a lot ! $\endgroup$
    – user15757055
    Sep 24, 2021 at 18:04

1 Answer 1

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The meaning of the RHS is $(x-y)$ times:

  • when $n=2$: $x+y$;
  • when $n=3$: $x^2+xy+y^2$;
  • when $n=4$: $x^3+x^2y+xy^2+xy^3$

and so on … In general you have $(x-y)$ times the sum of all expressions $x^iy^j$ with $i,j\in\Bbb N$ and $i+j=n-1$.

And if you multiply $(x-y)$ by that expression, for each $x^iy^j$ that you get, you'll also get a $-x^iy^j$. For instance:\begin{align}(x-y)(x^2+xy+y^2)&=x^3+x^2y+xy^2-x^2y-xy^2-y^3\\&=x^3-y^3.\end{align}

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  • $\begingroup$ How do you get theses results for n = 2,3 please ? For exemple with n = 3, when I do it myself I get : $x^{3-1}+x^{3-2}y+xy^{3-2}+y^{3-1}$ which is equal to $x^2 + 2xy + y^2$ because both $x^{3-2}y$ and $xy^{3-2}$ are equal to xy right ? Thanks a lot for the help $\endgroup$
    – user15757055
    Sep 23, 2021 at 14:12
  • $\begingroup$ What I get, when $n=3$, is$$x^{3-1}+x^{3-2}y^1+y^2=x^2+xy+y^2.$$ $\endgroup$
    – José Carlos Santos
    Sep 23, 2021 at 14:18
  • $\begingroup$ @user15757055 the $x^{3-2}y$ and the $xy^{3-2}$ terms are the same term. They only appear in the sum once, not twice. The dots do not directly imply that terms appearing before the dots are separate than the terms which appear after the dots, but rather the dots imply that there is some pattern that the terms in the sum follow. With few enough terms, the number of actual terms may be fewer than the number of pictured terms. $\endgroup$
    – JMoravitz
    Sep 23, 2021 at 14:32
  • $\begingroup$ @JoséCarlosSantos Did you mean $x^iy^j$ such that $i+j=n-1$? $\endgroup$
    – Lau
    Sep 23, 2021 at 14:42
  • $\begingroup$ @Lau Yes! I've edited my answer. Thank you. $\endgroup$
    – José Carlos Santos
    Sep 23, 2021 at 15:02

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