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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.

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  • $\begingroup$ Use induction on n $\endgroup$
    – Navin
    Commented Oct 14, 2016 at 22:27
  • $\begingroup$ Please verify the edit. $\endgroup$
    – Em.
    Commented Oct 14, 2016 at 22:27
  • $\begingroup$ Btw you are missing a y. $\endgroup$
    – MrYouMath
    Commented Oct 14, 2016 at 22:28
  • $\begingroup$ Now this is probably silly, but if you wanted to, you could consider the polynomial $f(x,y)=x^n-y^n$, which has obvious root/factor $(x-y)$, so upon dividing, you can use Viete's formulas to find all the other coefficients. $\endgroup$ Commented Oct 14, 2016 at 23:05
  • $\begingroup$ See also this question and other posts linked there. $\endgroup$ Commented Oct 15, 2016 at 0:46

5 Answers 5

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I like using $\sum$ like this:

$\begin{array}\\ (x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1}) &=(x-y)\sum_{k=0}^{n-1} x^{n-1-k} y^k\\ &=x\sum_{k=0}^{n-1} x^{n-1-k} y^k-y\sum_{k=0}^{n-1} x^{n-1-k} y^k\\ &=\sum_{k=0}^{n-1} x^{n-k} y^k-\sum_{k=0}^{n-1} x^{n-1-k} y^{k+1}\\ &=\sum_{k=0}^{n-1} x^{n-k} y^k-\sum_{k=1}^{n} x^{n-k} y^{k}\\ &=x^n+\sum_{k=1}^{n-1} x^{n-k} y^k-\left(y^n+\sum_{k=1}^{n-1} x^{n-k} y^{k}\right)\\ &=x^n-y^n+\sum_{k=1}^{n-1} x^{n-k} y^k-\sum_{k=1}^{n-1} x^{n-k} y^{k}\\ &=x^n-y^n\\ \end{array} $

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You have an error; it should be $x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})$. Just distribute.

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  • $\begingroup$ Honestly more comment worthy. And already mentioned in the comments. $\endgroup$ Commented Oct 14, 2016 at 23:04
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The simplest way is to prove first $$1-t^n=(1-t)(1+t+\dots+t^{n-1})$$ by induction.

The formula is trivial for $n=1$. So suppose the formula is valid for some $n\ge 1$, and consider the exponent $n+1$. \begin{align} 1-t^{n+1}&=(1-t^n)+(t^n-t^{n+1})\\ &=(1-t)(1+t+\dots+t^{n-1})+t^n(1-t)\\ &=(1-t)(1+t+\dots+t^{n-1+t^n}), \end{align} which proves the inductive step.

Now for the general formula: set $y=tx$, and substitute in the expression: \begin{align} x^n-y^n&=x^n(1-t^n)=x^n(1-t)(1+t+\dots+t^{n-1})\\ &=x(1-t)\cdot x^{n-1}(1+t+\dots+t^{n-1})\\ &=(x-tx)(x^{n-1}+x^{n-2}tx+\dots +(tx)^{n-1})\\ &=(x-y)(x^{n-1}+x^{n-2}y+\dots +y^{n-1}). \end{align}

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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*}

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  • $\begingroup$ Thanks very much @markuscheuer. This was really helpful. $\endgroup$
    – od320
    Commented Oct 25, 2016 at 15:37
  • $\begingroup$ @od320: You're welcome! $\endgroup$ Commented Oct 25, 2016 at 15:46
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Equation 1 by distributive law and Equation 2 also by distributive law (just expand the expressions)

Another method: Use polynomial long division on $$(x^n-y^n):(x-y)$$.

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