The correct option is A n is not a multiple of 3 and odd number
Let
f(x)=(x+1)n−xn−1
As x3+x2+x=x(x2+x+1)=x(x−ω)(x−ω2) is factor of f(x),
⇒f(0)=0,f(ω)=0,f(ω2)=0
Now (i)f(0)=(0+1)n−0n−1=0
f(0)=0 ∀n∈Z
(ii)f(ω)=(ω+1)n−ωn−1=(−ω2)n−ωn−1
Now, f(ω)=0 iff n is not multiple of 3 and odd number, as (−ω2)n−ωn−1=−(ω2n+ωn+1)=0
(iii)f(ω2)=(ω2+1)n−ω2n−1=(−ω)n−ω2n−1)
Now, f(ω2)=0 iff n is not multiple of 3 and odd number, as (−ω)n−ω2n−1=−(ωn+ω2n+1)=0
∴n should not be a multiple of 3 and n is odd number.