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How does;

$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$

work on $x^{2} - y^{2}$

When I attempt to apply the formula on $x^{2} - y^{2}$

I get the following

$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$

$x^{2} - y^{2} =(x-y)(2x+2y)$

which is obviously false. What is the correct way to use the formula?

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Try rewriting it in the proper notation instead of dots. – user2345215 Mar 1 '14 at 16:23
wouldnt we then enter the $x^{n-3}$ term, but how would that help? – user129967 Mar 1 '14 at 16:25
You have to stop when the power of x becomes 0. – kmitov Mar 1 '14 at 16:26
Thats what I've done ,haven't I? – user129967 Mar 1 '14 at 16:26
Regarding your edit , that is an interesting question that should be asked separately , in a separate thread. First check to see if it has been asked and answered already. :) – neofoxmulder Mar 1 '14 at 16:57

4 Answers 4

up vote 2 down vote accepted

You have too many terms, $x^{n-n}y^{n-1}$ is the last, which in this case is your second term $x^{n-2}y$. The third term would in your case be $x^{n-3}y^{n-2}$ which we don't want since $n=2$ and the exponent of $x$ would become negative.

To more easily see this, write the factorization in this form:

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

In this case it becomes

$$x^n - y^n = (x-y)(x^1y^0 + x^0y^1) = (x-y)(x + y)$$

which of course is the correct factorization.

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Thanks, I found your 1st equation very useful. I can see what you mean now. – user129967 Mar 1 '14 at 16:32

$$x^2 - y^2 = (x -y)(x^1y^0 + x^0y^1) =(x-y)(x+y)$$

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If you use proper notation, instead of dots, you get the correct result:


Then $x^2-y^2=(x-y)(x^{1-0}y^0+x^{1-1}y^1)=(x-y)(x+y).$

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Note how the last term in the factorization of $x^n-y^n$ has $x$ raised to the $0$th power. That means you have to stop when $x$ is raised to the power of $0$. $$x^2-y^2=(x-y)(x^1y^0+x^0y^1)$$ $$\boxed{x^2-y^2=(x-y)(x+y)}$$

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