How do you factor $z^6-64p^6$ ?

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Answer 1 · 2016-03-10T23:01:19

$z^6-64p^6=(z-2p)(z^2+2pz+4p^2)(z+2p)(z^2-2pz+4p^2)$

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

Hence:

$z^6-64p^6$

$=(z^3)^2-(8p^3)^2$

$=(z^3-8p^3)(z^3+8p^3)$

$=(z^3-(2p)^3)(z^3+(2p)^3)$

$=(z-2p)(z^2+2pz+4p^2)(z+2p)(z^2-2pz+4p^2)$

How do you factor z^6-64p^6z^6-64p^6?