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(B) The given equation is $x^{2}+x+1=0$.Since the roots are $\alpha$ and $\beta$,we know that $\alpha + \beta = -1$ and $\alpha \beta = 1$.We need to

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$-1$

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(B) The given equation is $x^{2}+x+1=0$.Since the roots are $\alpha$ and $\beta$,we know that $\alpha + \beta = -1$ and $\alpha \beta = 1$.We need to find $\alpha^{2}+\beta^{2}$.Using the identity $\alpha^{2}+\beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta$,we substitute the values:$\alpha^{2}+\beta^{2} = (-1)^{2} - 2(1) = 1 - 2 = -1$.Alternatively,the roots of $x^{2}+x+1=0$ are the complex cube roots of unity,$\omega$ and $\omega^{2}$.Thus,$\alpha^{2}+\beta^{2} = \omega^{2} + (\omega^{2})^{2} = \omega^{2} + \omega^{4} = \omega^{2} + \omega = -1$ (since $1+\omega+\omega^{2}=0$).

If $\alpha$ and $\beta$ are roots of the equation $x^{2}+x+1=0$,then $\alpha^{2}+\beta^{2}$ is

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