How do you factor $x^15 - 1/64$ ?

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Answer 1 · 2016-04-28T08:59:54

$x^15-1/64=(x^5-1/4)(x^(10)+1/4x^5+1/16)$

$x^15-1/64=(x^5)^3-(1/4)^3$ and hence using identity

$a^3-b^3=(a-b)(a^2+ab+b^2)$ , we get

$(x^5)^3-(1/4)^3=(x^5-1/4)((x^5)^2+1/4x^5+(1/4)^2)$

= $(x^5-1/4)(x^(10)+1/4x^5+1/16)$

How do you factor x^15 - 1/64x^15 - 1/64?