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

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Answer 1 · 2018-03-30T10:57:40

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

We are looking at the sum of two squares

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

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

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

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

How do you factor g4−1g^4-1?