If roots of an equation `x^n-1=0a r e1,a_1,_2, a_(n-1),` then the value of `(1-a_1)(1-a_2)(1-a_3)(1-a_(n-1))` will be `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]