Prove that $ x^n - y^n = (x-y) (x^{n-1}+x^{n-2}y\,+ \,\,...\,\,+ y^{n-1})$

Question

Prove that $ x^n - y^n = (x-y). (x^{n-1}+x^{n-2}y\,+ \,\,...\,\,+ y^{n-1}) $; $\,\,\,\,\,$$x,y \in \mathbb{R}$

Answer 1

One should not use "$\ldots$" in strict formal reasoning, so you should first of all agree that the second parentheses expression can be written as $$ \sum_{k=1}^{n} x^{n-k}y^{k-1}$$ (for some nitpickers: under the convention that $x^0=y^0=1$). Then note that $$ x\cdot \sum_{k=1}^{n} x^{n-k}y^{k-1}=\sum_{k=1}^{n} x^{n+1-k}y^{k-1}=x^{n}+\sum_{k=2}^{n} x^{n+1-k}y^{k-1}=x^n+ \sum_{k=1}^{n-1} x^{n-k}y^{k}$$ and $$ y\cdot \sum_{k=1}^{n} x^{n-k}y^{k-1}=\sum_{k=1}^{n} x^{n-k}y^{k}= \sum_{k=1}^{n-1} x^{n-k}y^{k}+y^n$$ hence the difference is indeed $$(x-y)\sum_{k=1}^{n} x^{n-k}y^{k-1} = x^n-y^n. $$

Answer 2

Hints:

1. Use induction or directly prove that for each $n$, $a\in \Bbb R$, $$a^{n+1} - 1 = (a - 1)(1 + a + \cdots + a^n).$$
2. If $y=0$ the identity holds trivially. If no, valuate the above identity at $a = x / y$.

Answer 3

You might find it easier to prove the special case when $y=1$, that is, $x^n-1 = (x-1)(x^{n-1}+x^{n-2} + \cdots + x + 1)$. Accomplishing this, can you see how to prove the general case as a corollary?