Factorize Difference of Two Powers

This article teaches how to think in order to derive the factorization of the difference of powers without memorization.

Your ability to factorize $a^n - b^n$ will come in handy for a great variety of mathematical problems. It’s enough to retain that $a^2 - b^2 = (a-b)(a+b)$. For $n = 3$, the patterns complexify.

Just like the case $n = 2$, factorize $a - b$ as your first move:

$$(a-b)(?)$$

In the second bracket, we’re sure to have $a^2$ and $b^2$ so that at the end we obtain the terms $a^3$ and $b^3$. It’s a non-negotiable:

$$(a-b)(a^2 + b^2)$$

Observe what we have when we expand the previous expression:

$$a^3 + ab^2 - a^2b - b^3$$

The factor $a^2 + b^2$ introduced $a^3$ and $b^3$ as desired, as well as two extraneous terms. We need $ab^2 - a^2b$ to disappear. Adding $ab$ in the second bracket is exactly what you need for the sum to telescope:

$$(a-b)(a^2 + ab + b^2) = a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3$$

Let’s go to $n = 4$. You know the drill already! Factorize $a - b$ and add the non-negotiable terms:

$$(a-b)(a^3 + b^3)$$

We know that those non-negotiables will produce extraneous terms, and our job will be to add sufficiently many terms so that the middle sum telescopes.

$$a^4 + ab^3 - a^3b - b^4$$

Good candidates for this are $a^2b$ and $ab^2$:

$$(a-b)(a^3 + a^2b + ab^2 + b^3)$$

Expanding confirms the telescope:

$$(a-b)(a^3 + a^2b + ab^2 + b^3) = a^4 + a^3b + a^2b^2 + ab^3 - a^3b - a^2b^2 - ab^3 - b^4 = a^4 - b^4$$

The general pattern that the terms in the second bracket follow is:

$$\sum_{k=0}^{n-1} a^{n-1-k}\, b^k$$

Using that summation to generate the second bracket for $n = 5$, where $n - 1 = 4$:

$$(a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$$

If ever you don’t remember the summation, we explored an almost empirical method to determine what the terms of the second bracket must be.