How do you factor $g^4-1$ ?

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Answer 1 · 2018-03-30T10:57:40

$(g+1)(g-1)(g^2+1)$

We are looking at the sum of two squares

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

So applying that rule we get
$(g^2-1)(g^2+1)$

We can also see that the $(g^2-1)$ term is also a sum of two squares so it now looks like

$(g+1)(g-1)(g^2+1)$

How do you factor g^4-1g^4-1?