<s>[INST] How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$ I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation: 
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-1})$$
The solution to the problem, contained in the answer book, is as follows:
\begin{align*}
x^n-y^n &= (x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&= x(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&\qquad -[y(x^{n-1}+x^{n-2}y+{...}+xy^{n-2}+y^{n-1})]&\Rightarrow \mathbf{Equation1}\\
&=x^n+x^{n-1}y+\dotsb+x^2y^{n-2}+xy^{n-1}\\
&\qquad -[x^{n-1}y+x^{n-2}y^2+xy^{n-1}+y^n]&\Rightarrow \mathbf{Equation2}\\
&=x^n-y^n
\end{align*}
While I believe that the distributive law was used to arrive at Equation 1, I do not understand how Equation 2 was arrived at. 
I have tried to solve this independently to no avail. I cannot seem to understand how $x^n$ and $y^n$ came about in Equation 2 for example.
To summarise the question, what principle was Equation 2 based upon? And how was this applied in the above problem. 
 [/INST] 
The step from equation 1 to equation 2 uses both the distributive law as well as the commutative law.

In order to better see what's going on, we look at small special cases $n=2,3$. We also use somewhat more detailed transformations.

n=2:
  \begin{align*}
(x-y)\left(x^{1}+y^1\right)&=x\left(x^{1}+y^1\right)-y\left(x^1+y^1\right)\\
&=\left(x\cdot x^1+x\cdot y^1\right)-\left(y\cdot x^1+y\cdot y^1\right)\\
&=\left(x^2+xy\right)-\left(yx+y^2\right)\\
&=x^2+xy-yx-y^2\\
&=x^2-y^2
\end{align*}
n=3:
  \begin{align*}
&(x-y)\left(x^{2}+x^1y^1+y^2\right)\\
&\qquad=x\left(x^{2}+x^1y^1+y^2\right)-y\left(x^{2}+x^1y^1+y^2\right)\\
&\qquad=\left(x\cdot x^2+x\cdot x^1y^1+x\cdot y^2\right)-\left(y\cdot x^2+y\cdot x^1y^1+y\cdot y^2\right)\\
&\qquad=\left(x^3+x^2y+xy^2\right)-\left(yx^2+y^2x+y^3\right)\\
&\qquad=x^3+x^2y+xy^2-yx^2-y^2x-y^3\\
&\qquad=x^3-y^3
\end{align*}
general n:
  \begin{align*}
&(x-y)\left(x^{n-1}+x^{n-2}y^1+\cdots+xy^{n-1}+y^{n-1}\right)\\
&\qquad=x\left(x^{n-1}+x^{n-2}y^1+\cdots\cdots\cdot\cdot+xy^{n-1}+y^{n-1}\right)\\
&\qquad\qquad\qquad\ \ -y\left(x^{n-1}+x^{n-2}y^1+\cdots\cdots+xy^{n-1}+y^{n-1}\right)\\
&\qquad=\left(x\cdot x^{n-1}+\color{blue}{x\cdot x^{n-2}y^1}+\cdots\cdots+x\cdot xy^{n-1}+\color{blue}{x\cdot y^{n-1}}\right)\\
&\qquad\qquad\qquad\quad-\left(\color{red}{y\cdot x^{n-1}}+y\cdot x^{n-2}y^1+\cdots\cdots\cdot+\color{red}{y\cdot xy^{n-1}}+y\cdot y^{n-1}\right)\\
&\qquad=\left(x^{n}+ \color{blue}{x^{n-1}y}+\cdots\cdots\cdots\cdots+x^2y^{n-1}+\color{blue}{xy^{n-1}}\right)\\
&\qquad\qquad\ -\left( \color{red}{y x^{n-1}}+x^{n-2}y^2+\cdots\cdots\cdots\cdots\cdot+\color{red}{xy^{n-1}}+y^{n}\right)\\
&\qquad=x^{n}\;+ \color{blue}{x^{n-1}y}+\cdots\cdots\cdots\cdots+x^2y^{n-1}+\color{blue}{xy^{n-1}}\\
&\qquad\qquad\ \ -\color{red}{x^{n-1}y}-x^{n-2}y^2-\cdots\cdots\cdots\cdots-\color{red}{xy^{n-1}}-y^{n}\\
&\qquad=x^n-y^n
\end{align*}

</s>