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

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Answer 1 · 2016-07-26T16:04:51

Factoring

$f(y) = y^(8n) - 2y^(4n) + 1 = (y^(4n) - 1)^2 =$
$= (y^(2n) + 1)^2(y^(2n) - 1)^2 = (y^(2n) + 1)^2(y^n + 1)^2(y^n -1)^2$

NOTE. We may continue, if required, to factor $(y^n - 1)^2$

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