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How to divide it

$${\frac {{x}^{n-2}-{y}^{n-2}}{x-y}}$$

to remove the $x-y$ term from the denominator.

We may assume that $n>2$ is an integer.

Thanks.

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1  
Do you know how to simply $\frac {x^k - y^k}{x-y}$? If so, let $k = n-2$. –  Calvin Lin Dec 29 '12 at 16:42
    
Thanks for the comment Calvin. We want to remove (x-y) from denom... –  Mia Dec 29 '12 at 16:44

6 Answers 6

up vote 4 down vote accepted

Hint: $$\alpha^n-\beta^n=(\alpha-\beta)(\alpha^{n-1}+\alpha^{n-2}\beta+...+\alpha\beta^{n-2}+\beta^{n-1})$$

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Hint:
$1-q^k=(1-q)\left(1+q+q^2+\ldots+q^{k-1} \right)$

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$${\frac {{x}^{n-2}-{y}^{n-2}}{x-y}}={(x-y)(x^{n-3}+x^{n-4}y+...+xy^{n-4}+y^{n-3})\over x-y}=x^{n-3}+x^{n-4}y+...+xy^{n-4}+y^{n-3}$$

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$x^{n-2}\cdot x \neq x^{n-2}$ –  Matema Tika Dec 29 '12 at 17:10
    
Yes I correct it –  Adi Dani Dec 29 '12 at 17:13

If you really want to you binomial coefficient as you wrote in the tag, you can expand everything around $x-y$, i.e. write $x^{n-2}-y^{n-2} = (y+ (x-y))^{n-2}-y^{n-2}$. The term $y^{n-2}$ cancels and then you can elegantly divide by $x-y$, as all the rest of terms are divisible by it.

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$\frac{x^{n-2}-y^{n-2}}{x-y}=\frac{(x-y)(x^{n-3}+\cdots y^{n-3})}{x-y}=x^{n-3}+\cdots y^{n-3}$

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To do this by direct division, as suggested in the question, use polynomial division (treating this as a polynomial in $x$) and note: $$x^{n-2}-y^{n-2}=(x-y)x^{n-3}+x^{n-3}y-y^{n-2}$$

then we could note: $$= (x-y)x^{n-3}+(x^{n-3}-y^{n-3})y$$

so that there is the possibility of an induction by applying a result for a lower index to the second term, or simply proceed alternatively:

$$=(x-y)x^{n-3}+(x-y)x^{n-4}y+x^{n-4}y^2-y^{n-2} = (x-y)(x^{n-3}+x^{n-4}y)+x^{n-4}y^2-y^{n-2}$$

and keep on dividing through, reducing the power of $x$ in the remainder each time until there is only a term in $y$ left - and discover that this cancels.

The factorisation, once discovered, comes up often enough to be worth remembering. There is a slightly different one for $\cfrac{x^n+y^n}{x+y}$ when $n$ is odd, which you might also like to find.

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