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If w is an imaginary cube root of unity ,then find the value of (1+w)(1+w^2)(1+w^3)(1+w^4)(1+w^5).................(1+w^3n)
Dear student w is cube root of unity so, w^3 =1 1+w+w^2 = 0 (1+w)(1+w^2) = 1+ w+ w^2 + w^3 w^3 = 1(1+w)(1+w^2)(1+w^3)(1+w^4)(1+w^5).................(1+w^3n)(1+w)(1+w^2)(1+w^3)(1+w^4)(1+w^5)(1+w^6)=> 2*2 We have n pairs of same brackets so , Ans = 2* 2* 2.......n times => 2^n
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