How do you factor completely 64-z^664−z6?
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)64−z6=(2−z)(4+2z+z2)(2+z)(4−2z+z2)
Explanation:
The difference of squares identity may be written:
a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)
The difference of cubes identity may 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 may be written:
a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)
So we find:
64-z^664−z6
=8^2-(z^3)^2=82−(z3)2
=(8-z^3)(8+z^3)=(8−z3)(8+z3)
=(2^3-z^3)(2^3+z^3)=(23−z3)(23+z3)
=(2-z)(2^2+2z+z^2)(2+z)(2^2-2z+z^2)=(2−z)(22+2z+z2)(2+z)(22−2z+z2)
=(2-z)(4+2z+z^2)(2+z)(4-2z+z^2)=(2−z)(4+2z+z2)(2+z)(4−2z+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)=(2−z)(2−ωz)(2−ω2z)(2+z)(2+ωz)(2+ω2z)
or if you prefer:
=(2-z)(2omega-z)(2omega^2-z)(2+z)(2omega+z)(2omega^2+z)=(2−z)(2ω−z)(2ω2−z)(2+z)(2ω+z)(2ω2+z)
where