• # question_answer Let $\alpha ,{{\alpha }^{2}}$be the roots of${{x}^{2}}+x+1=0$, then the equation whose roots are ${{\alpha }^{31}},{{\alpha }^{62}}$ is A) ${{x}^{2}}-x+1=0$   B) ${{x}^{2}}+x-1=0$ C) ${{x}^{2}}+x+1=0$ D) none of these

[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$