A) \[{{x}^{2}}-101x+100=0\]
B) \[{{x}^{2}}-102x+101=0\]
C) \[{{x}^{2}}+101x+100=0\]
D) \[{{x}^{2}}+102x+101=0\]
Correct Answer: B
Solution :
\[\because \,\,\,\,\,\,\alpha =w\] (complete non real cube of unity) \[\Rightarrow \,\,\,\,\,a=\left( 1+w \right)\left( 1+{{w}^{2}}+{{w}^{4}}+{{w}^{6}}+\text{ }.......\text{ }{{w}^{200}} \right)\] \[\Rightarrow \,\,\,\,\,a=\left( 1+w \right)\frac{\left( 1-{{\left( {{w}^{2}} \right)}^{101}} \right)}{\left( 1-{{w}^{2}} \right)}\] \[=\frac{\left( 1-w \right)\left( 1+w \right)}{1-{{w}^{2}}}=1\] And \[b=1+{{w}^{3}}+{{w}^{6}}+......\text{ }{{w}^{300}}=101\] Equation \[{{x}^{2}}-\left( 102 \right)x+101=0\]
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