Q)If $\alpha$ is a root of the equation $x^{2}+x+1=0$ and $(1+\alpha)^{7}=A+B\alpha+C(\alpha)^{2},$ then find the value of $5(3A-2B-C)$
Ans) My Approach:
I considered $\alpha=\omega$. Because we know that $\omega^{3}=1$. Therefore, $\omega^{2}+\omega+1=0$. Now $(1+\alpha)^{7}=(1+\omega)^{7}=(-\omega^{2})^{7}=-\omega^{14}$. Now we can write $\omega^{14}=(\omega^{12})(\omega^{2})$. Now $\omega^{12}=1$. Therefore, $-\omega^{14}=-\omega^{2}$.
Therefore, $A=0, B=0, C=-1$ . Therefore, $5(3A-2B-C)=5$ .
Please confirm whether my answer is correct or wrong.
As a hint: With respect to the equation $1+x+x^2=0 \to 1+\alpha+\alpha^2=0$ $$(1+\alpha)^{2}=1+2\alpha +\alpha^2=\underbrace{1+\alpha+\alpha^2}_{=0} +\alpha \\(1+\alpha)^2=\alpha$$ now $$(1+\alpha)^{4}=(\alpha)^{2}$$ you can multiply $$(1+\alpha)^{1}(1+\alpha)^{2}(1+\alpha)^{4}=(1+\alpha)^{7}\\=(1+\alpha)^{1}(\alpha)(\alpha)^{2}\\=(\alpha+\alpha^2)(\alpha)^{2}\\=-1(\alpha)^{2}$$