Let α ,β be the roots of the equation x^2-√6x+3=0 such that Im(α )>Im(β ). Let a,b be integers not divisible by 3and n be a natural number such that…

Given,

${x}^{2}-\sqrt{6}x+3=0$x2−6​x+3=0 has roots \alpha &\beta

Now, solving ${x}^{2}-\sqrt{6}x+3=0$x2−6​x+3=0 we get,

$$x=\frac{\sqrt{6}\pm i\sqrt{6}}{2}=\sqrt{3}(\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i)$$x=26​±i6​​=3​(2​1​±2​1​i)

Now, taking $\alpha =\sqrt{3}({e}^{i\frac{\pi }{4}}),\beta =\sqrt{3}({e}^{-i\frac{\pi }{4}})$α=3​(ei4π​),β=3​(e−i4π​)

Now, solving $\frac{{\alpha }^{99}}{\beta }+{\alpha }^{98}={\alpha }^{98}(\frac{\alpha }{\beta }+1)$βα99​+α98=α98(βα​+1)

$$=\frac{{\alpha }^{98}(\alpha +\beta )}{\beta }$$=βα98(α+β)​

$$=\frac{{3}^{49}{({e}^{i\frac{\pi }{4}})}^{98}(\sqrt{6})}{\sqrt{3}({e}^{-i\frac{\pi }{4}})}$$=3​(e−i4π​)349(ei4π​)98(6​)​

$$={3}^{49}({e}^{i99\frac{\pi }{4}})\times \sqrt{2}$$=349(ei994π​)×2​

$$={3}^{49}(\mathrm{cos}\frac{99\pi }{4}+i\mathrm{sin}\frac{99\pi }{4})\times \sqrt{2}$$=349(cos499π​+isin499π​)×2​

$$={3}^{49}(\mathrm{cos}(25\pi -\frac{\pi }{4})+i\mathrm{sin}(25\pi -\frac{\pi }{4}))\times \sqrt{2}$$=349(cos(25π−4π​)+isin(25π−4π​))×2​

$$={3}^{49}(-1+i)$$=349(−1+i)

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

So, on comparing we get,

$$n=49,a=-1,b=1$$n=49,a=−1,b=1

$$\therefore n+a+b=49-1+1=49$$∴n+a+b=49−1+1=49