Confused about answer to Spivak's calculus in prologue I have attached the question and answer to this question from Spivak's Calculus.
The proposition we are supposed to prove:

$$
x^n - y^n = (x - y)\, 
(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})
$$

The proof of this proposition:

\begin{align}
(x - y) \, 
(x^{n-1} + x^{n-2}y &+ \dots + xy^{n-2} + y^{n-1}) \\[2pt] 
&= x \, (x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1}) \\ 
&\qquad{} - y \, (x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1}) \\[2pt]
&= \bigl[ x^n + x^{n-1}y + \dots + x^2y^{n-2} + xy^{n-1} \bigr] \\ 
&\qquad{} - \bigl[ x^{n-1}y + x^{n-2}y^2 + \dots + xy^{n-1} + y^n  \bigr] \\[2pt]
&= x^n - y^n
\end{align}

I do not understand how $x^2\,y^{n-2}$ in the third to last line cancels with $x^{n-2}\,y^2$ in the second to last line (after the negative sign is distributed to $x^{n-2}\,y^2$).
Also, I do not understand how what is within the ellipses is treated. do we conceive of what is in the ellipses as multiplied by $x$ and $-y$ when $x$ and $-y$ are distributed to the expression with which $(x-y)$ is multiplied originally? If so, how can the ellipses cancel with each other as they are multiplied by different variables which we don't know are equal?
thanks, let me know if my question lacks clarity and I will explain my questions better/further.

You can avoid the $\cdots$ which is confusing you.
You have to use alternatives.
We want to Prove :
$ x^{n}-y^{n} = (x-y)(x^{n-1}y^{0}+x^{n-2}y^{1}+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+x^{1}y^{n-2}+x^{0}y^{n-1}) $
Write it alternately like this :
$$Z = (x-y)(x^{n-1}y^{0}+x^{n-2}y^{1}+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+x^{1}y^{n-2}+x^{0}y^{n-1})$$
$$Z = (x-y)[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m}]$$
(You can verify that this $ \Sigma $ matches Exactly what we want)
$$Z = (x[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m}]-y[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m}])$$
$$Z = ([\Sigma_{m=0}^{m=n-1} x^{n-1-m+1}y^{m}]-[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m+1}])$$
$$Z = (x^{n}y^{0}+[\Sigma_{m=1}^{m=n-1} x^{n-1-m+1}y^{m}]-[\Sigma_{m=0}^{m=n-2} x^{n-1-m}y^{m+1}]-x^{0}y^{n})$$
( here I have taken out the first & the last terms from the two $ \Sigma $ terms )
$$Z = (x^{n}y^{0}-x^{0}y^{n})+[\Sigma_{m=1}^{m=n-1} x^{n-1-m+1}y^{m}]-[\Sigma_{m=0}^{m=n-2} x^{n-1-m}y^{m+1}]$$
We can see that that 2 $ \Sigma $ terms will cancel ; To make it Explicit , we can change the variables to $ i = m-1 $ ( or $ m = i+1 $ )  in the first $ \Sigma $ term & to $ i = m $ ( or $ m = i $ ) in the second $ \Sigma $ term.
$$0 = [\Sigma_{i=1}^{i=n-2} x^{n-1-i-1+1}y^{i+1}]-[\Sigma_{i=0}^{i=n-2} x^{n-1-i}y^{i+1}]$$
$$0 = [\Sigma_{i=1}^{i=n-2} x^{n-1-i}y^{i+1}]-[\Sigma_{i=0}^{i=n-2} x^{n-1-i}y^{i+1}]$$
$$0 = [\Sigma_{i=1}^{i=n-2} ( x^{n-1-i}y^{i+1} - x^{n-1-i}y^{i+1} )]$$
$$0 = [\Sigma_{i=1}^{i=n-2} ( 0 )]$$
Hence :
$$Z = (x^{n}y^{0}-x^{0}y^{n})+[0]$$
$$(x^{n}-y^{n}) = (x-y)(x^{n-1}y^{0}+x^{n-2}y^{1}+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+x^{1}y^{n-2}+x^{0}y^{n-1})$$
