How do you factor completely #64-z^6#?
1 Answer
Use difference of squares, difference of cubes and sum of cubes identities to find:
#64-z^6 = (2-z)(4+2z+z^2)(2+z)(4-2z+z^2)#
Explanation:
The difference of squares identity may be written:
#a^2-b^2=(a-b)(a+b)#
The difference of cubes identity may be written:
#a^3-b^3=(a-b)(a^2+ab+b^2)#
The sum of cubes identity may be written:
#a^3+b^3=(a+b)(a^2-ab+b^2)#
So we find:
#64-z^6#
#=8^2-(z^3)^2#
#=(8-z^3)(8+z^3)#
#=(2^3-z^3)(2^3+z^3)#
#=(2-z)(2^2+2z+z^2)(2+z)(2^2-2z+z^2)#
#=(2-z)(4+2z+z^2)(2+z)(4-2z+z^2)#
The remaining quadratic factors have no simpler factors with Real coefficients, but if you allow Complex coefficients then you can go a little further:
#=(2-z)(2-omega z)(2-omega^2 z)(2+z)(2+omega z)(2+omega^2 z)#
or if you prefer:
#=(2-z)(2omega-z)(2omega^2-z)(2+z)(2omega+z)(2omega^2+z)#
where
