Difference of powers ($a^n-b^n=(a-b){\sum }_{k=0}^{n-1}{a}^{n-1-k}{b}^k$)

For $n\ge 1$ and $a,b$ in a commutative ring, multiplying out the displayed polynomial factorization cancels all intermediate terms, leaving $a^n-b^n$. The case $n=2$ is the difference of squares. Unlike division by $a-b$, this identity also applies when $a=b$.

For real $1\le b\le a$, every summand lies between one and $a^{n-1}$. Consequently $n(a-b)\le a^n-b^n\le n(a-b)a^{n-1}$. This turns a one-step discrepancy into a bound after repeated multiplication, for example when estimating the global error of a one-step method on $y^{\prime }=y$.