A) 0
B) 1
C) \[n\]
D) \[{{n}^{2}}\]
Correct Answer: C
Solution :
Since \[1,{{a}_{1}},{{a}_{2}}......,{{a}_{n-1}}\] are the roots of\[{{x}^{n}}-1=0\] \[\therefore \] \[{{x}^{n}}-1=(x-1)(x-{{a}_{1}})(x-{{a}_{2}})....\] \[(x-{{a}_{n-1}})\] \[\Rightarrow \]\[\frac{{{x}^{n}}-1}{x-1}=(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n-1}})\] \[\Rightarrow \]\[(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n-1}})\] \[={{x}^{n-1}}+{{x}^{n-2}}+....+{{x}^{2}}+x+1\] On putting\[x=1\]in the above equation, we get \[(1-{{a}_{1}})(1-{{a}_{2}})...(1-{{a}_{n-1}})\] \[=1+1+....+1+1+1\] \[\Rightarrow \] \[(1-{{a}_{1}})(1-{{a}_{2}})....(1-{{a}_{n-1}})=n.\] Note: Sum of roots of unity is always zero.
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