If $\alpha$ satisfies the equation $x^2+x+1=0$ and $(1+\alpha)^7=A+B\alpha+C\alpha^2$, $A,B,C\ge 0$, then $5(3A-2B-C)$ is equal to ______
Step-by-Step Solution
Key Concept: Roots of $x^2+x+1=0$ are $\omega,\omega^2$ (cube roots of unity). Take $\alpha=\omega$. Use $1+\omega=-\omega^2$, so $(1+\omega)^7=(-\omega^2)^7=-\omega^{14}=-\omega^2=1+\omega$. Then read off $A,B,C$.
$\alpha=\omega$, $1+\omega=-\omega^2$, $(1+\omega)^7=(-\omega^2)^7=-\omega^{14}=-\omega^2=1+\omega$.
So $A=1, B=1, C=0$.
$5(3A-2B-C)=5(3-2-0)=5$.
Correct Answer: 5