Question

How do you factor r^8-1

Answer 1

lemme just add a little bit to the superb reply by @Lammetthash

1 = 1

1² = 1

1³ = 1

1⁸⁹ = 1

1¹⁰⁰⁰⁰⁰⁰⁰⁰⁰ = 1

therefore

$\bf r^8-1\implies r^8-1^8\implies r^{4\cdot 2}-1^{4\cdot 2}\implies (r^4)^2-(1^4)^2 \\\\\\ \begin{array}{rrrr} (r^4-1^4)&&(r^4+1^4)\\[1em] [(r^2)^2-(1^2)^2]\\\\ (r^2-1^2)&(r^2+1^2)\\[1em] (r-1)(r+1)\\ \cline{1-3} \end{array}\\\\(r-1)(r+1)(r^2+1)(r^4+1)$

Answer 2

There's a simple factorization for the difference of squares:

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

Here, we have

$r^8-1=(r^4)^2-1^2=(r^4-1)(r^4+1)$

But $r^4$ is another square, so we can do more:

$r^4-1=(r^2)^2-1^2=(r^2-1)(r^2+1)$

Not done yet!

$r^2-1=(r-1)(r+1)$

Putting everything together, we have

$r^8-1=(r-1)(r+1)(r^2+1)(r^4+1)$