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