# Sum and difference of powers

The sum and difference of powers are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.

## Sums of Powers $a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})$

## Differences of Powers

If p is a positive integer and x and y are real numbers, $x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)$

For example, $x^2-y^2=(x-y)(x+y)$ $x^3-y^3=(x-y)(x^2+xy+y^2)$ $x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$

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

For example, $(y+1)^2-y^2=(y+1)+y$ $(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2$ $(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3$

If we also know that $y\geq 0$ then $2y\leq (y+1)^2-y^2\leq 2(y+1)$ $3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2$ $4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3$ $(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p$