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

2 Answers
Jun 12, 2015

Answer:

#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

Answer:

#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).#