Factoring differences of powers

The situation becomes more complicated when we look at power functions where the power is 2 or greater. To help with the analysis, recall that a difference of powers may always be factored. In particular, if p is a positive integer and x and y are real numbers,  

$$x^{p+1} - y^{p+1} = (x-y)(x^{p} + x^{p-1}y + \cdots + xy^{p-1} + y^{p}).$$ (1)

For example,

$$\begin{array} {rcl} x^2 - y^2 & = & (x-y)(x+y)\\ x^3 - y^3 &... ...^2)\\ x^4 - y^4 & = & (x-y)(x^3 + x^2y + xy^2 + y^3)\end{array}$$

Observe that the number of terms in the ``long'' factor is equal to the exponent in the expression being factored.

An amazing thing happens when x and y differ by 1, say, x = y+1. Then x-y = 1 and

$$\begin{array} {rcl} x^{p+1} - y^{p+1} & = & (y+1)^{p+1} - y... ...^{p} + (y+1)^{p-1}y + \cdots + (y+1)y^{p-1} + y^{p}.\end{array}$$

For example,

$$\begin{array} {rcl} (y+1)^2 - y^2 & = & (y+1) + y\\ (y+1)^3 ... ...)^4 - y^4 & = & (y+1)^3 + (y+1)^2y + (y+1)y^2 + y^3)\end{array}$$

If we also know that $y \geq 0$ then

$$\begin{array} {rcl} 2y \;\leq\; (y+1)^2 - y^2 \;\leq\; 2(y+1... ...\leq \;(y+1)^{p+1} - y^{p+1}\; \leq\; (p+1)(y+1)^{p}\end{array}$$