How do you factor completely #x^6-1#?

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Answer 1 · 2016-05-13T11:58:19

#(x-1)(x+1)(x^2-x+1)(x^2+x+1)#
#=(x-1)(x+1)(x-1/2-i sqrt3/2)(x-1/2+i sqrt3/2)(x+1/2-i sqrt3/2)(x+1/2+ i sqrt3/2)#

The answer in the given form can be obtained directly by observing that it is a cubic in #x^2#.

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

#=(x^2-1)((x^2)^2+x^2+!)#

#=(x-1)(x+1)(x^4+x^2+1)#

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

#=(x-1)(x+1)(x-1/2-i sqrt3/2)(x-1/2+i sqrt3/2)#

#(x+1/2-i sqrt3/2)(x+1/2+ isqrt3/2)#

The expanded form can be directly obtained from

#(x^6-1)=prod(x-e^(i(2kpi)/6))#, k=0, 1, 2,..,5.,

using the six 6th roots of 1, #{e^(i(2kpi)/6)}#, k=0, 1, 2,..,5.,