How do you factor z^6 - 1?

2 Answers
Jun 12, 2015

z^6-1 = (z^3-1)(z^3+1)
= (z-1)(z^2+z+1)(z+1)(z^2-z+1)

Explanation:

z^6-1 = (z^3)^2-1^2 is a difference of squares, so we can use the identity:

a^2-b^2 = (a-b)(a+b) to get:

(z^3)^2-1^2 = (z^3-1)(z^3+1)

z^3-1 = z^3-1^3 is a difference of cubes, so we can use the identity:

a^3-b^3=(a-b)(a^2+ab+b^2) to get:

z^3-1^3 = (z-1)(z^2+z+1)

z^3+1 = z^3 + 1^3 is a sum of cubes, so we can use the identity:

a^3+b^3=(a+b)(a^2-ab+b^2) to get:

z^3+1^3=(z+1)(z^2-z+1)

Putting this all together we get:

z^6-1 = (z-1)(z^2+z+1)(z+1)(z^2-z+1)

If we allow complex coefficients then we can factor:

z^6-1 = (z-1)(z-omega)(z-omega^2)(z+1)(z+omega)(z+omega^2)

where omega = -1/2+sqrt(3)/2i is the primitive cube root of unity.

Jun 12, 2015

z^6-1=(z-1)(z^2+z+1)(z+1)(z^2-z+1).

Explanation:

Remembering the formula a^2-b^2=(a-b)(a+b) we can state that z^6-1=(z^3-1)(z^3+1).

Now we have two ugly binomial to simplify. We could remember the formula of the difference of two perfect cubes , but it's not practical to memorize all formulas.

So what I suggest is to try to devide the two binomial by (z-1) and (z+1) that are the most probable divisors.
(z^3-1):(z+1)-> "quotient"=z^2-z+1 " remainder"=-2
(z^3-1):(z-1)-> " quotient"=z^2+z+1 " remainder"=0
So the second division is perfect!
(z^3+1):(z+1)-> " quotient"=z^2-z+1 " remainder"=0
(z^3+1):(z-1)-> "quotient"=z^2+z+1 " remainder"=2
And the first division is perfect!

Now we can rewrite the initial binomial in this way:
z^6-1=(z^3-1)(z^3+1)=(z-1)(z^2+z+1)(z^3+1)=(z-1)(z^2+z+1)(z+1)(z^2-z+1).