Given:
z=21−i3
⇒z=−(2−1+i3)
⇒z=−ω
where, ω is the cube root of unity. So,
1+ω+ω2=0
ω3=1
Now, let
A=(z+z1)3+(z2+z21)3+(z3+z31)3+…+(z21+z211)3
⇒A=(−ω−ω1)3+(ω2+ω21)3+(−ω3−ω31)3+….+(−ω21−ω211)3
⇒A=−(ωω2+1)3+(ω2ω4+1)3+(−1−11)3+….+(−ω21−ω211)3
⇒A=−(ω−ω)3+(ω2−ω2)3+(−1−1)3+….+(−ω21−ω211)3
⇒A=1+(−1)+(−1−1)3+1+(−1)+(−1−1)3….+(−1−1)3
⇒A=3[1+(−1)]+(−1−1)3
⇒A=−8
Then,
21+A=21−8
=13