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If $\alpha, \beta$ are the roots of $x^2-x+1=0$, then the quadratic equation whose roots are $\alpha^{2015}$, $\beta^{2015}$ is
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Verified Answer
The correct answer is:
$x^2+x+1=0$
We have, $x^2-x+1=0$
$\Rightarrow$ $x=\omega, \omega^2$
Then,
$\alpha=\omega$ and $\beta=\omega^2$
Now,
$$
\alpha^{2015}=\omega^{2015}=\omega^{3 \times 671+2}=\omega^2
$$
$\omega^2$ and $\beta^{2015}=\left(\omega^2\right)^{2015}=\omega^{4030}$
$$
=\omega^{3 \times 1343}+1=\omega
$$
$\therefore \quad \alpha^{2015}+\beta^{2015}=\omega^2+\omega=-1$
and $\quad \alpha^{2015} \cdot \beta^{2015}=\omega^2 \cdot \omega=\omega^3=1$
$\therefore$ Equation whose roots are $\alpha^{2015}$ and $\beta^{2015}$ will be
$$
\begin{aligned}
& x^2-(-1) x+1=0 \\
& \Rightarrow \quad x^2+x+1=0 \\
&
\end{aligned}
$$
$\Rightarrow$ $x=\omega, \omega^2$
Then,
$\alpha=\omega$ and $\beta=\omega^2$
Now,
$$
\alpha^{2015}=\omega^{2015}=\omega^{3 \times 671+2}=\omega^2
$$
$\omega^2$ and $\beta^{2015}=\left(\omega^2\right)^{2015}=\omega^{4030}$
$$
=\omega^{3 \times 1343}+1=\omega
$$
$\therefore \quad \alpha^{2015}+\beta^{2015}=\omega^2+\omega=-1$
and $\quad \alpha^{2015} \cdot \beta^{2015}=\omega^2 \cdot \omega=\omega^3=1$
$\therefore$ Equation whose roots are $\alpha^{2015}$ and $\beta^{2015}$ will be
$$
\begin{aligned}
& x^2-(-1) x+1=0 \\
& \Rightarrow \quad x^2+x+1=0 \\
&
\end{aligned}
$$
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