If a and b are real numbers such that (2 + α) ^ 4 = a + bα , where...

$\left(2 + \alpha\right){}^{4} = a + b\alpha$
${\left(4 + {\alpha}^{2} + 4\alpha\right)}^{2} = a + b\alpha$

Now using, $\alpha = \dfrac{- 1 + i\sqrt{3}}{2} ⇒ {\left(2\alpha + 1\right)}^{2} = {\left(i\sqrt{3}\right)}^{2}$

$⇒ {\alpha}^{2} = - \left(1 + \alpha\right)$

So, $a + b\alpha = {\left(4 - \alpha - 1 + 4\alpha\right)}^{2}$

$a + b\alpha = {\left(3 + 3\alpha\right)}^{2} = 9\left(1 + 2\alpha + {\alpha}^{2}\right)$
$a + b\alpha = 9\left(1 + 2\alpha + \left(- 1 - \alpha\right)\right) = 9\alpha$

$⇒ a = 0, b = 9 ⇒ a + b = 9$