Let alpha be a root of the equation 1+x^{2}+x | vedclass

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(A) The given equation is $x^{4}+x^{2}+1=0$.This can be factored as $(x^{2}+x+1)(x^{2}-x+1)=0$.The roots of $x^{2}+x+1=0$ are $\omega, \omega^{2}$ and

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(A) The given equation is $x^{4}+x^{2}+1=0$.This can be factored as $(x^{2}+x+1)(x^{2}-x+1)=0$.The roots of $x^{2}+x+1=0$ are $\omega, \omega^{2}$ and the roots of $x^{2}-x+1=0$ are $-\omega, -\omega^{2}$,where $\omega = e^{i2\pi/3}$.In all cases,$\alpha^{6}=1$ because $\alpha^{2}$ is a root of $y^{2}+y+1=0$,so $(\alpha^{2})^{3}=1$.We calculate the powers:$\alpha^{1011} = (\alpha^{6})^{168} \cdot \alpha^{3} = 1^{168} \cdot \alpha^{3} = \alpha^{3}$.Since $\alpha^{2}+\alpha^{4}+1=0$,we have $\alpha^{2}+\alpha^{4} = -1$. Also $\alpha^{6}=1 \Rightarrow \alpha^{3} = \pm 1$.For $x^{4}+x^{2}+1=0$,if $\alpha^{2} = \omega$,then $\alpha^{3} = \pm \sqrt{\omega} = \pm e^{i\pi/3}$.Actually,$\alpha^{6}=1$ and $\alpha^{2} 
eq 1$. Thus $\alpha^{3} = \pm 1$.Specifically,$\alpha^{1011} = (\alpha^{3})^{337} = (\pm 1)^{337} = \pm 1$.$\alpha^{2022} = (\alpha^{6})^{337} = 1^{337} = 1$.$\alpha^{3033} = (\alpha^{6})^{505} \cdot \alpha^{3} = 1 \cdot \alpha^{3} = \pm 1$.Thus,$\alpha^{1011}+\alpha^{2022}-\alpha^{3033} = \pm 1 + 1 - (\pm 1) = 1$.

Let $\alpha$ be a root of the equation $1+x^{2}+x^{4}=0$. Then the value of $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}$ is equal to

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