z3+2z2+2z+1=0 ⇒ (z+1)(z2+z+1)=0 z+1=0orz2+z+1=0 when z=−1,z1985+z100+1=(−1)1985+(−1)100+1 =−1+1+1=1≠0 ∵ z=−1 do not satisfy the equation z1985+z100+1=0 Then, it is not the common roots. ∴ The common roots are the roots of z2+z+1=0 We know that ω and ω2 are the roots of z2+z+1=0 (where ω is cube root of unity) When z=ω, (ω)1985+(ω)100+1=(ω3)661ω2+(ω3)33ω+1 =ω2+ω+1=0 (true) When z=ω, (ω2)1985+(ω2)100+1=(ω)3970+ω200+1 =(ω2)1323ω+(ω3)66ω2+1 =(1)1323ω+(1)66ω2+1 =ω+ω2+1=0 (true) ∴ ω,ω2 are the common roots of z1985+z100+1=0 and z3+2z2+2z+1=0.