Question:
If $\mathrm{a}$ and $\mathrm{b}$ 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: 4,
Solution:
$\alpha=\omega$
$\left(\omega^{3}=1\right)$
$\Rightarrow \quad(2+\omega)^{4}=a+b \omega$
$\Rightarrow \quad 2^{4}+4 \cdot 2^{3} \omega+6 \cdot 2^{2} \omega^{3}+4 \cdot 2 \cdot \omega^{3}+\omega^{4}$
$=a+b \omega$
$\Rightarrow \quad 16+32 \omega+24 \omega^{2}+8+\omega=a+b \omega$
$\Rightarrow \quad 24+24 \omega^{2}+33 \omega=a+b \omega$
$\Rightarrow \quad-24 \omega+33 \omega=a+b \omega$
$\Rightarrow \quad a=0, b=9$
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