Solution
$\mathrm{x}^{2}-\sqrt{6} \mathrm{x}+3=0 \Rightarrow \mathrm{x}=\frac{\sqrt{6}}{2}(1 \pm \mathrm{i})=\frac{\sqrt{3}(1 \pm \mathrm{i})}{\sqrt{2}}$. Roots are $\alpha=\sqrt{3}\left(\mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}\right)$
\[
\because \frac{\alpha^{99}}{\beta}+\alpha^{98}=\frac{\alpha^{98}(\alpha+\beta)}{\beta}=\frac{\left(\sqrt{3} \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right)^{98}\left[\sqrt{3} \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}+\sqrt{3} \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}\right]}{\sqrt{3} \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}}=3^{49} \mathrm{e}^{\mathrm{i} \frac{49 \pi}{2}}\left(2 \cos \frac{\pi}{4}\right) \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}=3^{49} \frac{(\mathrm{i})}{\sqrt{2}} 2\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)=3^{49}(-1+\mathrm{i})=3^{\mathrm{n}}(\mathrm{a}+\mathrm{ib})
\]
we get $\mathrm{n}=49, \mathrm{a}=-1, \mathrm{~b}=1 \therefore \mathrm{n}+\mathrm{a}+\mathrm{b}=49-1+1=49$
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