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

1 Answer
Dec 20, 2015

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 #omega = -1/2 + sqrt(3)/2 i# is the primitive Complex cube root of #1#