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

1 Answer
Mar 10, 2016

Answer:

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

Explanation:

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)#