How do you factor completely #x^9-1#?

2 Answers
Apr 6, 2016

Answer:

#x^9-1=(x-1)(x^2+x+1)(x^6+x^3+1)#

Explanation:

To factorize #x^9-1#, we can use the identity

#a^3-b^3=(a-b)(a^2+ab+b^2)#.

Hence #x^9-1=(x^3)^3-1^3=(x^3-1)(x^6+x^3+1)#

factorizing #x^3-1# further

#x^9-1=(x^3-1)(x^6+x^3+1)#

= #(x-1)(x^2+x+1)(x^6+x^3+1)#

Apr 6, 2016

Answer:

The separate factors are #(x-(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8#. The first factor #(x-1)# is the only real factor.

Explanation:

The roots of this equation #x^9-1=0# are the 9 values of #x=1^(1/9)#.
They are#(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8#.
k=0 gives the only real root 1.

So, the factors of #x^0-1# are #(x-(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8#.