Let ω be the cube root of unity.
(1+ω)n=nC0+nC1ω+nC2ω2+....
Since, ω3n=1, ω3n+1=ω, ω3n+2=ω2, n=0,1,2,...
⇒(1+ω)n=(nC0+nC3+...)+(nC1+nC4+...)(−1+√3i2)+(nC2+nC5+...)(−1−√3i2)
=(nC0+nC3+...)−12(nC1+nC2+nC4+nC5+....)+i√32(nC1−nC2+nC4−nC5+....)
⇒(1+ω)n=(a−b2)+i√3c2
Taking modulus both the side, we get
|(−ω2)n|=(a−b2)2+3c24 [∵1+ω+ω2=0]
⇒(a−b2)2+3c24=1