# How do you factor y^(8n) - 2y^(4n) +1?

$f \left(y\right) = {y}^{8 n} - 2 {y}^{4 n} + 1 = {\left({y}^{4 n} - 1\right)}^{2} =$
$= {\left({y}^{2 n} + 1\right)}^{2} {\left({y}^{2 n} - 1\right)}^{2} = {\left({y}^{2 n} + 1\right)}^{2} {\left({y}^{n} + 1\right)}^{2} {\left({y}^{n} - 1\right)}^{2}$
NOTE. We may continue, if required, to factor ${\left({y}^{n} - 1\right)}^{2}$