If $\alpha_1,\alpha_2,\ldots,\alpha_n$ be the roots of the equation $x^n+1$

Question

then $(1-\alpha_1)(1-\alpha_2)\ldots(1-\alpha_n)$ equals to ? I think here we need the info of whether $n$ is even or odd else how will we say whether by vieta's formula what is the value of $1+(-1)^n$ ?

Answer 1

If they are indeed the roots of $x^n+1$, then

$$(x-a_1)(x-a_2)(x-a_3)\dots(x-a_n)=x^n+1$$

So for $x=1$, the whole thing equals $2$

Answer 2

Let $1-x=y\iff x=?$

So, $\displaystyle0=x^n+1=(1-y)^n+1=(-y)^n+\binom n{n-1}(-y)^{n-1}+\cdots+1$

$\displaystyle\iff(-1)^ny^n+n(-1)^{n-1}y^{n-1}\cdots+2=0$

$\displaystyle(-1)^n\prod_{r=1}^ny_r=\dfrac2{(-1)^n}\iff\prod_{r=1}^ny_r=?$