<s>[INST] Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
          &= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
          &= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2}  - \cdots- x y^{n-1} - y^n \\
          &= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n  \\
          &\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:


*
  
*Associate law for addition
  
*Existence of an additive identity
  
*Existence of additive inverses
  
*Commutative law for additions
  
*Associative law for multiplication
  
*Existence of a multiplicative identity
  
*Existence of multiplicative inverses
  
*Commutative law for multiplication
  
*Distibutive law

 [/INST] Sorry but I don't speak English well but I will try to write as correctly as possible.
Your demonstration is correct, but your idea of cancellation is incorrect.
$x^n−y^n=(x−y)(x^\left(n−1\right)+x^\left(n−2\right)y+⋯+xy^\left(n−2\right)+y^\left(n−1\right))$
When we multiply first for 'x' to our terms of the second parenthesis let's notice that all terms that contains 'x' go up one grade, this by exponent laws.
When we multiplies for 'y' to our terms of the second parenthesis let's notice that all terms that contains y go up one grade but the terms that contains x remain the same.
Therefore at the moment to cancel our terms, "ellipsis" of the formula cancel the terms left over in your demostration.
This is your result  $x^n+x^2y^\left(n−2\right)−x^\left(n−2\right)y^2−y^n$ the extra term $+x^2y^\left(n−2\right)$is "cancel" by the "ellipsis" because the term that is not visible but is in the "ellipsis" in the second parenthesis is
$(x^\left(n−1\right)+x^\left(n−2\right)y+⋯ (x^2y^\left(n-3\right)) ⋯+xy^\left(n−2\right)+y^\left(n−1\right)$ because on the left one degree goes up the 'x' and one degree goes down the y and cancel your term. Analogously the other extra term $-x^\left(n−2\right)y^2$ is cancel by the other "ellipsis" that came out of the first multiplication by x.
$x^n−y^n=(x−y)(x^\left(n−1\right)+x^\left(n−2\right)y+⋯+xy^\left(n−2\right)+y^\left(n−1\right))$
Dem.
By distributive law we have
$x(x^\left(n-1\right)+x^\left(n-2\right)y+...+xy^\left(n-2\right)+y^\left(n-1\right))$
$-y(x^\left(n-1\right)+x^\left(n-2\right)y+...+xy^\left(n-2\right)+y^\left(n-1\right))$   from where

$\quad x^n+x^\left(n-1\right)y+.........+x^2y^\left(n-2\right)+xy^\left(n-1\right)$
$\qquad -(x^\left(n-1\right)y+x^\left(n-2\right)y^2+.........+xy^\left(n-1\right)+y^n)$ therefore
$x^n-y^n$
</s>