Given any sequence say (a0,a1,a2,a3,....) we represent in generating functions as
a0 + a1.z + a2.z2 + a3.z3 + a4.z4 + ....
where z is called as indicator variable.
Now given in question that
a0 + a1.z + a2.z2 + a3.z3 + a4.z4 + ....
= (1 + z) / (1 - z)3
= (1 + z).( 1 / (1-z)3 )
= (1 + z).(1 / 1-z)3
= (1 + z).(1 + z + z2 + z3 +z4 + ....)3
= 1.(1 + z + z2 + z3 +z4 + ....)3 + z.(1 + z + z2 + z3 +z4 + ....)3 ==> eqn 1
From eqn 1, I can say that a0 = coefficent of z0 and a3 = coefficient of z3
1.(1 + z + z2 + z3 +z4 + ....)3
= 1.(1 + z + z2 + z3 +z4 + ....).(1 + z + z2 + z3 +z4 + ....).(1 + z + z2 + z3 +z4 + ....)
coefficient of z0 = 1 (Because z0 is possible only if 1 is taken in all 3 bracket terms)
coefficient of z3 = number of ways of choosing (1,z,z2) + (1,1,z3) + (z,z,z)
= (3 * 2) + (3C2 ) + 1
= 6 + 3 + 1
= 10
z.(1 + z + z2 + z3 +z4 + ....)3
= z.(1 + z + z2 + z3 +z4 + ....).(1 + z + z2 + z3 +z4 + ....).(1 + z + z2 + z3 +z4 + ....)
coefficient of z0 = 0 (as we have a z outside and z0 is not possible.
coefficient of z3 = coefficient of z2 from bracket terms
= number of ways choosing (1,1,z2) + (1,z,z)
= 3C2 + 3C2
= 6
a0 = Total coefficient of z0 = 1 + 0 = 1
a3 = Total coefficent of z3 = 10 + 6
= 16
a3 - a0 = 16 - 1 = 15.