If a1,a2,....,an−1 are the nth roots of unity then the value of (1−a1)(1−a2)...(1−an−1) is equal to
A
√3
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B
12
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C
n
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D
0
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Solution
The correct option is Cn Since, 1,a1,a2,....,an−1 are the nth root of unity. ∴xn−1=(x−1)(x−a1)...(x−an−1) ⇒xn−1x−1=(x−a1)(x−a2)....(x−an−1) ∴xn−1+xn−2+...+x2+x+1 =(x−a1)(x−a2)....(x−an−1) Put x=1, we get (1−a1)(1−a2)...(1−an−t)=1+1+...+ntimes =n.