How do you factor completely 64-z^664z6?

1 Answer
Dec 20, 2015

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)64z6=(2z)(4+2z+z2)(2+z)(42z+z2)

Explanation:

The difference of squares identity may be written:

a^2-b^2=(a-b)(a+b)a2b2=(ab)(a+b)

The difference of cubes identity may be written:

a^3-b^3=(a-b)(a^2+ab+b^2)a3b3=(ab)(a2+ab+b2)

The sum of cubes identity may be written:

a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2ab+b2)

So we find:

64-z^664z6

=8^2-(z^3)^2=82(z3)2

=(8-z^3)(8+z^3)=(8z3)(8+z3)

=(2^3-z^3)(2^3+z^3)=(23z3)(23+z3)

=(2-z)(2^2+2z+z^2)(2+z)(2^2-2z+z^2)=(2z)(22+2z+z2)(2+z)(222z+z2)

=(2-z)(4+2z+z^2)(2+z)(4-2z+z^2)=(2z)(4+2z+z2)(2+z)(42z+z2)

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)=(2z)(2ωz)(2ω2z)(2+z)(2+ωz)(2+ω2z)

or if you prefer:

=(2-z)(2omega-z)(2omega^2-z)(2+z)(2omega+z)(2omega^2+z)=(2z)(2ωz)(2ω2z)(2+z)(2ω+z)(2ω2+z)

where omega = -1/2 + sqrt(3)/2 iω=12+32i is the primitive Complex cube root of 11