Multiple choice

If roots of an equation $x^n-1=0$ are $1,a_1,a_2....a_{n-1}$, then the value of $(1-a_1)(1-a_2)(1-a_3).....(1-a_{n-1})$ will be

  1. $n$
  2. $n^2$
  3. $n!$
  4. $0$
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A Correct answer
Explanation

The polynomial x^n - 1 can be factored as (x-1)(x-a1)(x-a2)...(x-an-1). Dividing by (x-1) gives the sum of geometric series 1 + x + x^2 + ... + x^(n-1) = (x-a1)(x-a2)...(x-an-1). Setting x=1 gives 1 + 1 + ... + 1 (n times) = n.

AI explanation

The roots of x to the power of n minus 1 equal 0 are the n-th roots of unity, and they satisfy the polynomial factorization x to the power of n minus 1 equals the quantity x minus 1 times the quantity x minus a sub 1 times the quantity x minus a sub 2 and so on up to the quantity x minus a sub n minus 1. Dividing both sides by x minus 1 gives the limit as x approaches 1 on the left side, which equals the derivative of x to the power of n minus 1 evaluated at 1, yielding n. The right side becomes the quantity 1 minus a sub 1 times the quantity 1 minus a sub 2 times the quantity 1 minus a sub 3 and so on up to the quantity 1 minus a sub n minus 1, so the product equals n.

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