If $ \alpha_1,\alpha_n, ....\alpha_n $ be the roots of the equation $x^n+1=0$, then $(1-\alpha_1)(1-\alpha_2)...(1-\alpha_n)$ is equal to a) 1. b) 0 c)n d)2
when I put n=3,and directly evalutae the value of $(1-\alpha_1)(1-\alpha_2)...(1-\alpha_n)$ I find no option matching to my answer.
I know the answer is 2. i've seen the solution here If $\alpha_1,\alpha_2,\ldots,\alpha_n$ be the roots of the equation $x^n+1$.
but, if we put n=3 , $x^3+1$ has roots -1 , and $\frac{1}{2} \pm \frac{3i}{2}$ then if we put that in the expression $(1-\alpha_1)(1-\alpha_2)...(1-\alpha_n)$ , it doesn't give 2. what mistake am i doing ?