# 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.

• Use induction on n Oct 14, 2016 at 22:27
– Em.
Oct 14, 2016 at 22:27
• Btw you are missing a y. Oct 14, 2016 at 22:28
• 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. Oct 14, 2016 at 23:05

I like using $\sum$ like this:

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

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.

• Honestly more comment worthy. And already mentioned in the comments. Oct 14, 2016 at 23:04

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}

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*}

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