Complex Number Question 5
Question 5 - 2024 (27 Jan Shift 1)
If $\alpha$ satisfies the equation $x^{2}+x+1=0$ and $(1+\alpha)^{7}=\mathrm{A}+\mathrm{B} \alpha+\mathrm{C}^{2}, \mathrm{~A}, \mathrm{~B}, \mathrm{C} \geq 0$, then
$5(3 \mathrm{~A}-2 \mathrm{~B}-\mathrm{C})$ is equal to
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Answer (5)
Solution
$x^{2}+x+1=0 \Rightarrow x=\omega, \omega^{2}=\alpha$
Let $\alpha=\omega$
Now $(1+\alpha)^{7}=-\omega^{14}=-\omega^{2}=1+\omega$
$\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0$
$\therefore 5(3 A-2 B-C)=5(3-2-0)=5$
