If a and b are real numbers such that (2+alph | vedclass

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(D) Given $\alpha = \frac{-1+i \sqrt{3}}{2} = \omega,$ where $\omega$ is the complex cube root of unity.We have $\omega^2 + \omega + 1 = 0$ and $\omeg

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$9$

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(D) Given $\alpha = \frac{-1+i \sqrt{3}}{2} = \omega,$ where $\omega$ is the complex cube root of unity.We have $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$.Expanding $(2+\omega)^4$ using the binomial theorem:$(2+\omega)^4 = 2^4 + 4(2^3)(\omega) + 6(2^2)(\omega^2) + 4(2)(\omega^3) + \omega^4$$= 16 + 32\omega + 24\omega^2 + 8(1) + \omega$$= 24 + 33\omega + 24\omega^2$Since $\omega^2 = -1 - \omega$,we substitute this into the expression:$= 24 + 33\omega + 24(-1 - \omega)$$= 24 + 33\omega - 24 - 24\omega$$= 9\omega$Comparing this with $a + b\omega$,we get $a = 0$ and $b = 9$.Therefore,$a + b = 0 + 9 = 9$.

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

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