Sum of the common roots of z2006+z100+1=0 and z3+2z2+2z+1=0 is
0
-1
1
2
(z3+1)+(2z2+2z)=0 ⇒(z+1)(z2−z+1)+2z(z+1)=0⇒(z+1)(z2+z+1)=0⇒z=−1,w,w2 Out of these w and w2 satisfy z2006+z100+1=0 ∴Sum of common roots =w+w2=−1