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If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+x+1=0$, then what is the equation whose roots are $\alpha^{19}$ and $\beta^{7}$ ?
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The correct answer is:
$x^{2}+x+1=0$
If $\alpha$ and $\beta$ are the roots of the equations $x^{2}+x+1=0$
$\Rightarrow \alpha=\omega$ and $\beta=\omega^{2}$
or, $\alpha=\omega^{2}$ and $\beta=\omega$
$\alpha^{19}+\beta^{7}=\omega^{19}+\omega^{14}=\omega+\omega^{2}=-1$
$\alpha^{19}+\beta^{7}=\omega^{38}+\omega^{7}=\omega^{2}+\omega=-1$
$+\beta^{7}=-1$
and $-\beta^{7}=\omega^{19} \cdot \omega^{14}=\omega^{33}=1$
or $\omega^{38} \cdot \omega^{7}=\omega^{45}=$
and $\beta$ The is
$x^{2}-\left(\alpha^{19}+\beta^{7}\right) x+\alpha^{19} \beta^{7}=0 \Rightarrow \quad x^{2}+x+1$
$=0$
$\Rightarrow \alpha=\omega$ and $\beta=\omega^{2}$
or, $\alpha=\omega^{2}$ and $\beta=\omega$
$\alpha^{19}+\beta^{7}=\omega^{19}+\omega^{14}=\omega+\omega^{2}=-1$
$\alpha^{19}+\beta^{7}=\omega^{38}+\omega^{7}=\omega^{2}+\omega=-1$
$+\beta^{7}=-1$
and $-\beta^{7}=\omega^{19} \cdot \omega^{14}=\omega^{33}=1$
or $\omega^{38} \cdot \omega^{7}=\omega^{45}=$
and $\beta$ The is
$x^{2}-\left(\alpha^{19}+\beta^{7}\right) x+\alpha^{19} \beta^{7}=0 \Rightarrow \quad x^{2}+x+1$
$=0$
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