Sum of coefficients of terms of odd powers of x in (1+x−x2−x3)8 is

Given expansion (1+x−x2−x3)8 is Let consider f(x)=(1+x−x2−x3)8  substitute x=1 then we get ⇒f(1)=(1+1−1−1)5=0    Similarly let consider f(x)=(1+x−x2−x3)8  substitute x=−1 then we get  ⇒f(−1)=(1−1−1+1)8=0 Sum of coefficients of odd powers of x=f(1)−f(−1)2=0−02                                                                                                                                                       =0