Let α, β be the roots of the equation x ^ 2 - 6 x + 3 = 0 such that...

Given,

${x}^{2} - \sqrt{6}x + 3 = 0$ has roots $\alpha \text{ and } \beta$

Now, solving ${x}^{2} - \sqrt{6}x + 3 = 0$ we get,

$x = \dfrac{\sqrt{6} \pm i\sqrt{6}}{2} = \sqrt{3}\left(\dfrac{1}{\sqrt{2}} \pm \dfrac{1}{\sqrt{2}}i\right)$

Now, taking $\alpha = \sqrt{3}\left({e}^{i\dfrac{\pi}{4}}\right), \beta = \sqrt{3}\left({e}^{- i\dfrac{\pi}{4}}\right)$

Now, solving $\dfrac{{\alpha}^{99}}{\beta} + {\alpha}^{98} = {\alpha}^{98}\left(\dfrac{\alpha}{\beta} + 1\right)$

$= \dfrac{{\alpha}^{98} ( \alpha + \beta )}{\beta}$

$= \dfrac{{3}^{49}{\left({e}^{i\dfrac{\pi}{4}}\right)}^{98}\left(\sqrt{6}\right)}{\sqrt{3}\left({e}^{- i\dfrac{\pi}{4}}\right)}$

$= {3}^{49}\left({e}^{i99\dfrac{\pi}{4}}\right) \times \sqrt{2}$

$= {3}^{49}\left(cos\dfrac{99\pi}{4} + isin\dfrac{99\pi}{4}\right) \times \sqrt{2}$

$= {3}^{49}\left(\cos \left(25\pi - \dfrac{\pi}{4}\right) + isin\left(25\pi - \dfrac{\pi}{4}\right)\right) \times \sqrt{2}$

$= {3}^{49} ( - 1 + i )$

$= {3}^{n} ( a + ib )$

So, on comparing we get,

$n = 49, a = - 1, b = 1$

$∴ n + a + b = 49 - 1 + 1 = 49$