The correct option is A −1
z3+2z2+2z+1=0
z3+z2+z2+z+z+1=0
(z3+z2+z)+(z2+z+1)=0
z(z2+z+1)+1(z2+z+1)=0
(z+1)(z2+z+1)=0
Thus
z=−1 and z=w,w2
Clearly z=−1 does not satisfy second equation.
w2015+w100+1=w2+w+1=0
w4030+w200+1=w+w2+1=0
Hence w,w2 are the two common roots of the above given equations.
Now,
w+w2
=−1 ....(since 1+w+w2=0)
Hence, option 'A' is correct.