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 $5(3 A-2 B-C)$ is equal to ____________.
Answer (integer)
5
Solution
<p>$x^2+x+1=0 \Rightarrow x=\omega, \omega^2=\alpha$</p>
<p>Let $\alpha=\omega$</p>
<p>Now $(1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega$</p>
<p>$$\begin{aligned}
& A=1, B=1, C=0 \\
& \therefore 5(3 A-2 B-C)=5(3-2-0)=5
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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