If $\alpha$ is the nth root of unity , then the value of

Question

Given , $\alpha$ is the nth root of unity . Then , the value of $$(11-\alpha)(11-\alpha^2)(11-\alpha^3)........(11-\alpha^{n-1})$$ is equal to ...

I tried to use the property : That sum of nth roots of unity is 0 and they form a G.P series. But I couldn't solve this question, Suggest some simpler methods or any hint to solve this . Thank you.

Answer 1

The roots of $x^n-1=0$ are $1,\alpha, \alpha ^2,...,\alpha ^{n-1}$. So $x^n-1=(x-1)(x-\alpha)(x-\alpha ^2)...(x-\alpha ^{n-1})$. Put $x=11$.