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Roots of Unity

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If $1,\alpha, \alpha^2,.....,\alpha^{n - 1}$ be the $n^{th}$ roots of unity, then $(1-\alpha)(1-\alpha^2).....(1-\alpha^{n-1}) $

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$3$

💡 Explanation:

Basically $1,a^1,a^2......a^{n-1}$ all these are the roots of this equation $x^3 – 1 =0$

So we can write

$x^3 -1= (x-1)(x-a _1)(x-a _2).....(x-a _{n-1})$

$\dfrac{x^3 -1}{(x-1)}= (x-a^1)(x-a^2).....(x-a^{n-1})$

$x^2 + x +1= (x-a^1)(x-a^2).....(x-a^{n-1})$

Put $x=1$ on both the sides now

Ans $=3$