If (1, ₁, ₂, ., _ n-1 ) are the (n )th roots of unity and (n ) is an even natural number, then ( (l+ ₁ ) (l+ ₂ ) . (l+ _ n-1 )= )

Step-by-step solution

Since, 1, ( ₁, ₂ . . . . ., _ n-1 ) are the (n )th roots of unity, then (x^n-1=(x-1) (x- ₁ ) (x- ₂ ) ) ( (x- _ n-1 ) ) ( (x- ₁ ) (x- ₂ ) (x- _ n-1 )= x^n-1 x-1 ) On putting (x=-1 ), we get ( aligned & (-1- ₁ ) (-1- ₂ ) . . . (-1- _ n-1 )= (-1)^n-1 -1-1 & (1+ ₁ ) (1+ ₂ ) . . . (1+ _ n-1 )=0 aligned ) [as (n ) is an even natural number] Hence, option (c) is correct.