The given equation z3+2z2+2z+1=0 can be rewritten as (z+1)(z2+z+1)=0.
Its roots are −1,ω and ω2
Let f(z)=z1985+z100+1
Putting z=−1,ω and ω2 respectively, we get f(−1)=(−1)1985+(−1)100+1=0
Therefore, −1 is not a root of the equation f(z)=0
Again, f(ω)=ω1985+ω100+1 =(ω3)661ω2+(ω3)33ω+1 =ω2+ω+1=0
Therefore, ω is a root of the equation f(z)=0.
Similarly, f(ω2)=0
Hence, ω and ω2 are the common roots.