How do you factor z^6-64p^6z6−64p6?
1 Answer
Mar 10, 2016
z^6-64p^6=(z-2p)(z^2+2pz+4p^2)(z+2p)(z^2-2pz+4p^2)z6−64p6=(z−2p)(z2+2pz+4p2)(z+2p)(z2−2pz+4p2)
Explanation:
The difference of squares identity can be written:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
The difference of cubes identity can be written:
a^3-b^3 = (a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)
The sum of cubes identity can be written:
a^3+b^3 = (a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)
Hence:
z^6-64p^6z6−64p6
=(z^3)^2-(8p^3)^2=(z3)2−(8p3)2
=(z^3-8p^3)(z^3+8p^3)=(z3−8p3)(z3+8p3)
=(z^3-(2p)^3)(z^3+(2p)^3)=(z3−(2p)3)(z3+(2p)3)
=(z-2p)(z^2+2pz+4p^2)(z+2p)(z^2-2pz+4p^2)=(z−2p)(z2+2pz+4p2)(z+2p)(z2−2pz+4p2)
