If $a$ and $b$ are real numbers such that $(2+\alpha)^{4}=a+b \alpha$, where $\alpha=\frac{-1+i \sqrt{3}}{2}$, then $a+b$ is equal to :
Correct Option: 1
Given that, $\alpha=\frac{-1+\sqrt{3} i}{2}=\omega$
$\therefore(2+\omega)^{4}=a+b \omega \Rightarrow\left(4+\omega^{2}+4 \omega\right)^{2}=a+b \omega$
$\Rightarrow\left(\omega^{2}+4(1+\omega)\right)^{2}=a+b \omega$
$\Rightarrow\left(\omega^{2}-4 \omega^{2}\right)^{2}=a+b \omega$
$\Rightarrow\left(-3 \omega^{2}\right)^{2}=a+b \omega \Rightarrow 9 \omega^{4}=a+b \omega$$\quad\left[\because 1+\omega=-\omega^{2}\right]$
$\Rightarrow 9 \omega=a+b \omega \quad\left(\because \omega^{3}=1\right)$
On comparing, $a=0, b=9$
$\Rightarrow a+b=0+9=9$
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