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Discriminant

If 1,α, α2,...

Question

If $1, α, α^{2}, . . . . ., α^{n - 1}$ be the $n^{t h}$ roots of unity, then $(1 - α) (1 - α^{2}) . . . . . (1 - α^{n - 1})$

A

3

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B

0

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C

n

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D

1

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Solution

The correct option is B $3$
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
$x^{3} - 1 = (x - 1) (x - a_{1}) (x - a_{2}) . . . . . (x - a_{n - 1})$

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

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

Put $x = 1$ on both the sides now

Ans $= 3$

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Similar questions

1, α_{1}, α_{2}, α_{3}, . . . . ., α_{n - 1}

are the nth roots of unity then show that

n = (1 - α_{1}) (1 - α_{2}) (1 - α_{3}) . . . . . (1 - α_{n - 1})

α_{1}

α_{2}

α_{n - 1}

are the nth roots of unity then

(1 - α_{1}) (1 - α_{2}) . . . (1 - α_{n - 1}) =

1, α, α^{2} . . . . . α^{n - 1}

are n roots of unity then ,

1. α . α^{2} . . . . α^{n - 1}

Q. If n is an odd positive integer and

I, α_{1}, α_{2}, . . . . α_{n - 1}

n, n^{t h}

roots of unity, then

(3 + α^{1}) (3 + α^{2}) . . . . (3 + α^{n - 1})

α_{0}, α_{1}, α_{2}, \dots, α_{n - 1}

n^{t h}

roots of unity, then

α_{k} = c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})

0 \leq k \leq n - 1

(x^{n} - 1) = (x - α_{0}) (x - α_{1}) \dots (x - α_{n - 1})

(1 + α_{0}) (1 + α_{1}) \dots (1 + α_{n - 1})

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