If roots of an equation xn−1=0 are 1,a1,a2,…,an−1, then the value of 1−a11−a21−a3............(1-an-1) will be

Clearly,xn−1=(x−1)x−a1x−a2…x−an−1⇒ xn−1x−1=x−a1x−a2…x−anLHS is the summation of series in GP⇒ 1+x+x2+…+xn−1=x−a1x−a2…x−an−1⇒n=1−a11−a2…1−an−1 [putting x=1]

If roots of an equation xn−1=0 are 1,a1,a2,…,an−1, then the value of 1−a11−a21−a3............(1-an-1) will be

Clearly,xn−1=(x−1)x−a1x−a2…x−an−1⇒ xn−1x−1=x−a1x−a2…x−anLHS is the summation of series in GP⇒ 1+x+x2+…+xn−1=x−a1x−a2…x−an−1⇒n=1−a11−a2…1−an−1 [putting x=1]

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$If roots of an equation x^{n} - 1 = 0 are 1, a_{1}, a_{2}, \dots, a_{n - 1}, then the value of (1 - a_{1}) (1 - a_{2}) (1 - a_{3}) {............(1-a}_{n - 1}) will be$

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$n$

$n^{2}$

$n^{n}$

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answer is A.

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Detailed Solution

Clearly, $x^{n} - 1 = (x - 1) (x - a_{1}) (x - a_{2}) \dots (x - a_{n - 1})$

$\begin{array}{l} \Rightarrow \frac{x^{n} - 1}{x - 1} = (x - a_{1}) (x - a_{2}) \dots (x - a_{n}) \\ L H S i s t h e s u m m a t i o n o f s e r i e s i n G P \\ \Rightarrow 1 + x + x^{2} + \dots + x^{n - 1} = (x - a_{1}) (x - a_{2}) \dots (x - a_{n - 1}) \end{array}$