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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.

Gary
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user15757055
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1 Answers1

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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}

José Carlos Santos
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  • 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 – user15757055 Sep 23 '21 at 14:12
  • What I get, when $n=3$, is$$x^{3-1}+x^{3-2}y^1+y^2=x^2+xy+y^2.$$ – José Carlos Santos Sep 23 '21 at 14:18
  • @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. – JMoravitz Sep 23 '21 at 14:32
  • @JoséCarlosSantos Did you mean $x^iy^j$ such that $i+j=n-1$? – Lau Sep 23 '21 at 14:42
  • @Lau Yes! I've edited my answer. Thank you. – José Carlos Santos Sep 23 '21 at 15:02
  • thanks a lot everyone i think i get it now – user15757055 Sep 23 '21 at 15:05