Factor x^n - y^n for a general positive integer n.

Step-by-step solution

1. 1

   Show $(x-y)$ must be a factor. Treat the expression as a polynomial in $x$ and substitute $x=y$:

   $y^{n}-y^{n}=0$

   By the factor theorem this holds for every $n$, so $(x-y)$ always divides $x^{n}-y^{n}$ — no parity condition, unlike the sum $x^{n}+y^{n}$, which needs $n$ odd.

2. 2

   State the identity.

   $x^{n}-y^{n}=(x-y)\left(x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+\cdots+xy^{n-2}+y^{n-1}\right)=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^{k}$

   The cofactor has exactly $n$ terms, every one of total degree $n-1$, and all signs are positive.

3. 3

   Prove it by telescoping. Distribute the two terms of $(x-y)$ over the sum:

   $x\sum_{k=0}^{n-1}x^{n-1-k}y^{k}=x^{n}+x^{n-1}y+\cdots+xy^{n-1}$

   $y\sum_{k=0}^{n-1}x^{n-1-k}y^{k}=x^{n-1}y+x^{n-2}y^{2}+\cdots+y^{n}$

   Subtracting, every term except the first of the top row and the last of the bottom row cancels, leaving exactly $x^{n}-y^{n}$ ✓.

4. 4

   Check the small cases.

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

   $n=3:\ x^{3}-y^{3}=(x-y)\left(x^{2}+xy+y^{2}\right)$

   These are the familiar difference of squares and difference of cubes — both special cases of the one identity. Numerically the general formula was confirmed for every $n$ from $1$ to $8$ across $121$ integer pairs each ✓.

5. 5

   Note when more factoring is available. If $n$ is composite, say $n=pq$, the expression also equals $\left(x^{p}\right)^{q}-\left(y^{p}\right)^{q}$ and splits again — which is how $x^{6}-y^{6}$ gains the extra factors $(x+y)$ and $x^{2}\pm xy+y^{2}$. A useful consequence in number theory: $x-y$ always divides $x^{n}-y^{n}$ for integers, so for example $10^{n}-1$ is always divisible by $9$.