If `1,alpha,alpha^(2),……….,alpha^(n-1)` are `n^(th)` root of unity, the value of `(3-alpha)(3-alpha^(2))(3-alpha^(3))……(3-alpha^(n-1))`, is

Options

A. n
B. 0
C. `(3n-1)/2`
D. `(3n+1)/2`

Answer

C

Explanation

We have,
`x^(n)-1 = (x-1)(x-alpha)(x-alpha^(2))…..(x-alpha^(n-1))`
`rArr (x^(n)-1)/(x-1) = (x-alpha)(x-alpha^(2))…..(x-alpha^(n-1))`
`rArr 1+x+x^(2)+…..+x^(n-1)=(x-alpha)(x-alpha^(2))…..(x-alpha^(n-1))`
Putting `x=3`, We get
`1+3+3^(2)+......+ 3^(n-1)=(3-alpha^(2))...(3-alpha^(n-1))`
`rArr (3-alpha)(3-alpha^(2))....(3-alpha^(n-1))=(3^(n)-1)/(3-1) = (3^(n)-1)/(2)`