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(B) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$. Let $\alpha = \omega$.Since $1+\omega+\omega^2=0$,we have $1+\omega = -\omega^2$.Then $(1

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$5$

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(B) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$. Let $\alpha = \omega$.Since $1+\omega+\omega^2=0$,we have $1+\omega = -\omega^2$.Then $(1+\alpha)^7 = (1+\omega)^7 = (-\omega^2)^7 = -\omega^{14} = -\omega^2$.Using $1+\omega+\omega^2=0$,we have $-\omega^2 = 1+\omega$.Comparing $1+\omega$ with $A+B\alpha+C\alpha^2 = A+B\omega+C\omega^2$,we get $A=1, B=1, C=0$.Thus,$5(3A-2B-C) = 5(3(1)-2(1)-0) = 5(3-2) = 5(1) = 5$.

If $\alpha$ satisfies the equation $x^2+x+1=0$ and $(1+\alpha)^7=A+B\alpha+C\alpha^2$,where $A, B, C \geq 0$,then $5(3A-2B-C)$ is equal to:

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