An obvious thought about factors

The quadratic expression $a^2-b^2$ factorises as $(a-b)(a+b)$.

Similarly, the expression $a^2+b^2$ factorises over the complex numbers as $(a-bi)(a+bi)$.

And while I’d always sort-of-known that $a^3-b^3=(a-b)(a^2+ab+b^2)$, I hadn’t quite in my head made it $(a-b)(a-\omega b)(a-{\omega }^{2}b)$, where ${\omega }^3=1$.

It’s odd: I’d happily factorise $x^3-1$ using $\omega$, but the link between it and $a^3-b^3$ hadn’t really jumped at me, despite writing a post about Eisenstein integers recently.

In general, then, $a^n-b^n$ factorises as ${\mathrm{\Pi }}_{k=0}^{n-1}(a-b{w}^k)$, where $w^n=1$.

Why am I mentioning it? Well, mainly to show that sometimes I miss connections that are right there in front of me. But also because it might make someone else go “oo! That’s neat.”