Question

How do you factor `z^6 - 1`?

Answer 1

`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.

Answer 2

`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).`