How do you factor $(x^{8}) - 16$ $(x^8)-16$ ?

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How do you factor $(x^{8}) - 16$ $(x^8)-16$ ?

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Answer 1 · 2015-11-04T20:33:53

It is

$x^{8} - 16 = {(x^{4})}^{2} - 4^{2} = (x^{4} - 4) \cdot (x^{4} + 4) = (x^{2} - 2) \cdot (x^{2} + 2) \cdot [{(x^{2} + 2)}^{2} - 4 x^{2}] = (x^{2} - 2) \cdot (x^{2} + 2) \cdot (x^{2} + 2 - 2 x) \cdot (x^{2} + 2 + 2 x)$ $x^8-16=(x^4)^2-4^2=(x^4-4)*(x^4+4)=(x^2-2)*(x^2+2)*[(x^2+2)^2-4x^2]=(x^2-2)*(x^2+2)*(x^2+2-2x)*(x^2+2+2x)$

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How do you factor $(x^{8}) - 16$ $(x^8)-16$ ?

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