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)#