put =ωandnotethat1+ω+ω2=0
where ω=−12+i√32andω2=−12−i√32
Δ=(a0+a3+a6+......)+ω(a1+a4+......)+ω(a2+a3......)
or Δ=Aω+Bω2+C=0
equating real and imaginary part of both sides
A−B2−C2=0√32(B−C)=0
B = C and hence A = B
A \, = \, B \, = \., C
Again putting x = 1 in the given relation . we get 3^ n = sum of the all the condition = A + B + C