(1-w+w^2) (1-w^2+w^4) (1-w^4+w^8) to 2n factors is a)2 b)2^2n c)2n d)2^n

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Vikas TU · Answer

Dear student The ans woud be (x-1)^2n w^1 = w w^2 = w w^4 = w w^8 = w^6+w^2 = w^2 2^n = 3n , 3n + 1 , 3n +2 1+w+w^2 = 0 => (x-1)(x-1)(x-1)......2n => (x-1)^2n is ans

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(1-w+w^2) (1-w^2+w^4) (1-w^4+w^8) to 2n factors is a)2 b)2^2n c)2n d)2^n

(1-w+w^2) (1-w^2+w^4) (1-w^4+w^8) to 2n factors is
a)2 b)2^2n c)2n d)2^n

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(1-w+w^2) (1-w^2+w^4) (1-w^4+w^8) to 2n factors is a)2 b)2^2n c)2n d)2^n

(1-w+w^2) (1-w^2+w^4) (1-w^4+w^8) to 2n factors is a)2 b)2^2n c)2n d)2^n

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(1-w+w^2) (1-w^2+w^4) (1-w^4+w^8) to 2n factors is
a)2 b)2^2n c)2n d)2^n