InterviewSolution
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InterviewSolution
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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 (a) `n` (b) `n^2` (c) `n^n` (d) `0`A. nB. `n^(2)`C. `n^(n)`D. 0 |
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Answer» Correct Answer - 1 Clearly, `x^(n) - 1= (x - 1)(x - a_(1)) (x - a_(2)) .. (x - a_(n - 1))` `rArr (x^(n) - 1)/(x - 1) = (x - a_(1))(x - a_(2)) .. (x - a_(n - 1)` `rArr 1 + x + x^(2) + .. + x^(n - 1) = (x - a_(1))(x - a_(2))..(x - a_(n - 1))` `rArr n = (1 - a_(1))(1 - a_(2))..(1 - a_(n - 1))` [putting x = 1] |
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