If satisfies the equation x^2+x+1=0 and (1+ )^7=A+B +C ^2,A,B,C 0, then 5(3A-2B-C) is equa…

Question

If $\alpha $ satisfies the equation ${x}^{2}+x+1=0$ and $(1+\alpha {)}^{7}=A+B\alpha +C{\alpha }^{2},A,B,C\geqslant 0$, then $5(3A-2B-C)$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Since $\alpha$ is a non-real cube root of unity, $1+\alpha=-\alpha^2$. So $(1+\alpha)^7=(-\alpha^2)^7=-\alpha^{14}=-\alpha^{12}\cdot\alpha^2=-\alpha^2$ (since $\alpha^3=1$).

So $A+B\alpha+C\alpha^2=-\alpha^2=0+0\cdot\alpha+(-1)\alpha^2$... but since we need $A,B,C\ge0$, we instead use $\alpha^2=-1-\alpha$: $-\alpha^2=1+\alpha$, giving $A=1,B=1,C=0$.

$5(3A-2B-C)=5(3-2-0)=5.$

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