If roots of an equation `x^n-1=0` are `1,a_1,a_2,........,a_(n-1)` then the value of `(1-a_1)(1-a_2)(1-a_3)........(1-a_(n-1))` will be (a) `n` (b) `n^2` (c) `n^n` (d) `0`

Options

A. n
B. `n^(2)`
C. `n^(n)`
D. 0

Answer

1

Explanation

Clearly,
`x^(n) - 1= (x - 1)(x - a_(1)) (x - a_(2)) .. (x - a_(n - 1))`
`rArr (x^(n) - 1)/(x - 1) = (x - a_(1))(x - a_(2)) .. (x - a_(n - 1)`
`rArr 1 + x + x^(2) + .. + x^(n - 1) = (x - a_(1))(x - a_(2))..(x - a_(n - 1))`
`rArr n = (1 - a_(1))(1 - a_(2))..(1 - a_(n - 1))` [putting x = 1]