Given,
{z}^{1985}+{z}^{100}+1=0&{z}^{3}+2{z}^{2}+2z+1=0
Now solving,
z3+2z2+2z+1=0
⇒(z+1)(z2−z+1)+2z(z+1)=0
⇒(z+1)(z2+z+1)=0
⇒z=−1,z=ω,ω2
Now putting z=ω in z1985+z100+1 we get,
⇒ω1985+ω100+1
⇒ω2+ω+1=0 asω3n=1
Also, z=ω2
⇒ω3970+ω200+1
⇒ω+ω2+1=0
Also, z=−1 will not satisfy the equation z1985+z100+1
Hence, there are two common root