If n is a natural number that is not a multiple of 3, prove that 1+w^n+w^(2n)=01+wn+w2n=0?

1 Answer
Aug 7, 2018

Please go through the Explanation.

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.