Concept:Cube roots of unity: 1, ω, and ω2, where ω=2−1+i3, ω2=2−1−i3, with properties 1+ω+ω2=0 and ω3=1.Explanation:First, factorize z3+2z2+2z+1=0.(z+1)(z2+z+1)=0.So the roots are z=−1, z=ω, and z=ω2.Now check which of these also satisfy z2017+z2018+1=0.Rewrite the equation as z2017(z+1)+1=0.For z=−1:(−1)2017(−1+1)+1=0+1=1=0.So −1 is not a common root.For z=ω:ω2017(ω+1)+1=(ω3)672⋅ω⋅(−ω2)+1=1672⋅ω⋅(−ω2)+1=−ω3+1=−1+1=0.Thus ω is a common root.For z=ω2:(ω2)2017(ω2+1)+1=((ω3)672⋅ω)2⋅(−ω)+1=(1672⋅ω)2⋅(−ω)+1=ω2⋅(−ω)+1=−ω3+1=−1+1=0.Thus ω2 is also a common root.Therefore, the common roots are ω and ω2.Answer:1,ω2 and −1,ω2 (Option D).