JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 29)
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
5
Explanation
$$x^2+x+1=0 \Rightarrow x=\omega, \omega^2=\alpha$$
Let $$\alpha=\omega$$
Now $$(1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega$$
$$\begin{aligned} & A=1, B=1, C=0 \\ & \therefore 5(3 A-2 B-C)=5(3-2-0)=5 \end{aligned}$$
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