Solution:
The equation is expressed as,
z3+2z2+2z+1=0
(z+1)(z2−z+1)+2z(z+1)=0
(z+1)(z2+z+1)=0
After solving the above expression, we get
z=−1 and z=ω,ω2
If, z=−1 then,
z2018+z2017+1=0
(−1)2018+(−1)2017+1=0
1−1+1=1≠0
If, z = ω then,
z2018+z2017+1=0
(ω)2018+(ω)2017+1=0
ω2+ω+1=0
If, z=ω2 then,
z2018+z2017+1=0
(ω2)2018+(ω2)2017+1=0
ω2+ω+1=0
Hence, the common roots are ω and ω2.
Consider the equation,
z4+z2+1=0
If, z=ω then,
ω2+ω+1=0
If, z=ω2 then,
ω2+ω+1=0
Thus, the equation z4+z2+1=0, satisfy the common roots.
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