Question

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. $x^{60}+x^{30}+1=0$

Answer 1

Answer: (C)

Given equation is $x^2+x+1=0$. Since, $\alpha ,{\alpha }^2$ are the roots of the equation.
$\therefore \alpha +{\alpha }^2=-1\dots (i)$
and ${\alpha }^3=1\dots (ii)$
Now, for the equation of roots are ${\alpha }^{31}$ and ${\alpha }^{62}$.
${\alpha }^{31}+{\alpha }^{62}={\alpha }^{31}\left(1+{\alpha }^{31}\right)$
$\Rightarrow {\alpha }^{31}+{\alpha }^{62}={\alpha }^{30}\alpha \left(1+{\alpha }^{30}\cdot \alpha \right)$
$\Rightarrow {\alpha }^{31}+{\alpha }^{62}={\left({\alpha }^3\right)}^{10}\cdot \left\{1+{\alpha }^{30}\cdot \alpha \right\}$
$\Rightarrow {\alpha }^{31}+{\alpha }^{62}={\left({\alpha }^3\right)}^{10}\cdot \alpha \left\{1+{\left({\alpha }^3\right)}^{10}\cdot \alpha \right\}$
$\Rightarrow {\alpha }^{31}+{\alpha }^{62}=\alpha (1+\alpha )$ [using Eq. (ii)]
$\Rightarrow {\alpha }^{31}+{\alpha }^{62}=-1$ [using Eq. $(i)]$
Again ${\alpha }^{31}\cdot {\alpha }^{62}={\alpha }^{93}$
$={\left({\alpha }^3\right)}^{31}=1$
$\therefore$ Required equation is
$x^2-\left({\alpha }^{31}+{\alpha }^{62}\right)x+{\alpha }^{31}\cdot {\alpha }^{62}=0$
$\Rightarrow x^2+x+1=0$