Given,
x2−6x+3=0 has roots \alpha &\beta
Now, solving x2−6x+3=0 we get,
x=26±i6=3(21±21i)
Now, taking α=3(ei4π),β=3(e−i4π)
Now, solving βα99+α98=α98(βα+1)
=βα98(α+β)
=3(e−i4π)349(ei4π)98(6)
=349(ei994π)×2
=349(cos499π+isin499π)×2
=349(cos(25π−4π)+isin(25π−4π))×2
=349(−1+i)
=3n(a+ib)
So, on comparing we get,
n=49,a=−1,b=1
∴n+a+b=49−1+1=49