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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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: |
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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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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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