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:

We are asked to evaluate $$(2+\alpha)^4$$ where $$\alpha=\dfrac{-1+i\sqrt3}{2}$$ and then write the result in the form $$a+b\alpha$$ with real numbers $$a$$ and $$b$$. Afterwards we will calculate $$a+b$$.

First recall that $$\alpha$$ is a non-real cube root of unity. Hence it satisfies the standard relations

$$\alpha^3 = 1$$   and   $$1+\alpha+\alpha^2 = 0.$$

From $$1+\alpha+\alpha^2=0$$ we obtain the very useful identity

$$\alpha^2 = -1-\alpha.$$

Now we expand $$(2+\alpha)^4$$ with the Binomial Theorem. The theorem states that for any numbers $$x$$ and $$y$$,

$$ (x+y)^4=\;^4C_0x^4+\;^4C_1x^3y+\;^4C_2x^2y^2+\;^4C_3xy^3+\;^4C_4y^4,$$

where the binomial coefficients for the fourth power are $$^4C_0=1,\;^4C_1=4,\;^4C_2=6,\;^4C_3=4,\;^4C_4=1.$$

Putting $$x=2$$ and $$y=\alpha$$, we get

$$ (2+\alpha)^4 = 2^4\;+\;4\cdot2^3\alpha\;+\;6\cdot2^2\alpha^2\;+\;4\cdot2\alpha^3\;+\;\alpha^4. $$

Compute each numerical power of 2 first:

$$2^4=16,\qquad2^3=8,\qquad2^2=4.$$

Substituting these values,

$$ (2+\alpha)^4 = 16 \;+\; 4\cdot8\,\alpha \;+\; 6\cdot4\,\alpha^2 \;+\; 4\cdot2\,\alpha^3 \;+\; \alpha^4, $$

which simplifies to

$$ (2+\alpha)^4 = 16 \;+\; 32\alpha \;+\; 24\alpha^2 \;+\; 8\alpha^3 \;+\; \alpha^4. $$

We now reduce all higher powers of $$\alpha$$ using the identities already stated.

First, $$\alpha^3=1,$$ so

$$8\alpha^3 = 8\cdot1 = 8.$$

Next, $$\alpha^2=-1-\alpha,$$ so

$$24\alpha^2 = 24(-1-\alpha) = -24 - 24\alpha.$$

Finally, $$\alpha^4 = \alpha\cdot\alpha^3 = \alpha\cdot1 = \alpha.$$

Substituting each of these back into the expansion:

$$ (2+\alpha)^4 = 16 \;+\; 32\alpha \;+\;(-24 - 24\alpha) \;+\; 8 \;+\; \alpha. $$

Now collect the constant (real) terms and the coefficients of $$\alpha$$ separately.

Constant terms: $$16 + (-24) + 8 = 0.$$

$$\alpha$$ terms: $$32\alpha + (-24\alpha) + \alpha = (32-24+1)\alpha = 9\alpha.$$

Hence we have obtained the very compact result

$$ (2+\alpha)^4 = 0 + 9\alpha = 9\alpha. $$

In other words, $$a=0$$ and $$b=9.$$ Therefore,

$$a+b = 0+9 = 9.$$

Hence, the correct answer is Option A.