A) \[{{x}^{2}}-x+1=0\]
B) \[{{x}^{2}}+x-1=0\]
C) \[{{x}^{2}}+x+1=0\]
D) none of these
Correct Answer: C
Solution :
[c] : Given equation is \[{{x}^{2}}+x+1=0\] \[\therefore \] \[\alpha +{{\alpha }^{2}}=-1\] ...(i) and a3 = 1 ...(ii) We have to find the equation whose roots are \[{{\alpha }^{13}}\] and \[{{\alpha }^{62}}\]. Now,\[{{\alpha }^{31}}+{{\alpha }^{62}}={{\alpha }^{31}}(1+{{\alpha }^{31}})\] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}={{\alpha }^{31}}.\alpha (1+{{\alpha }^{30}}.\alpha )\] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}={{({{\alpha }^{3}})}^{10}}.\alpha \{1+{{({{\alpha }^{3}})}^{10}}.\alpha \}\] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}=\alpha (1+\alpha )\] [From(ii)] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}=-1\] [From(i)] Again, \[{{\alpha }^{31}}.{{\alpha }^{62}}={{\alpha }^{93}}\Rightarrow {{\alpha }^{31}}.{{\alpha }^{62}}={{[{{\alpha }^{3}}]}^{31}}=1\] Required equation is\[{{x}^{2}}-({{\alpha }^{31}}+{{\alpha }^{62}})x+{{\alpha }^{31}}.{{\alpha }^{62}}=0\] \[\Rightarrow \]\[{{x}^{2}}+x+1=0\]
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