If n is a natural number that is not a multiple of 3, prove that #1+w^n+w^(2n)=0#?

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Answer 1 · 2018-08-07T18:13:11

Please go through the Explanation.

#w^(3n)-1=(w^3)^n-1-1=0..............(ast^1)#.

Factorising, #w^(3n)-1=(w^n)^3-1^3,#

#=(w^n-1){(w^n)^2+w^n*1+1^2}#,

#:. w^(3n)-1=(w^n-1)(w^(2n)+w^n+1)......(ast^2)#.

Combining #(ast^1) and (ast^2)#, we get,

# (w^n-1)(w^(2n)+w^n+1)=0#

# rArr w^n-1=0, or, w^(2n)+w^n+1=0#.

Since, #n!=3m, m in NN, w^n!=1#.

#:. w^(2n)+w^n+1=0, if, n!=3m, m in NN#.