The general case of #(x+y+z)^3# is easier to handle due to its symmetry in #x#, #y# and #z#. So solve that first, then substitute #z=1# ...

#(x+y+z)^3 = (x+y+z)(x+y+z)(x+y+z)#

The coefficient of the resulting #x^3# term will be #1#, since there's only one way of picking #x# from each of the #3# trinomials. By symmetry the coefficient of #y^3# and #z^3# is also #1#.

The coefficient of the resulting #x^2y# term will be #3#, since there are #3# trinomials to pick #y# from, then one way to pick #x# from both of the remaining trinomials. By symmetry, #3# is also the coefficient of #y^2z#, #z^2x#, #x^2z#, #y^2x# and #z^2y#.

The coefficient of the resulting #xyz# term will be #6#, since therre are #3# trinomials to pick #x# from, #2# remaining trinomials to pick #y# from, then only one remaining trinomial to pick #z# from.

Putting this all together:

#(x+y+z)^3#

#=x^3+y^3+z^3+3x^2y+3y^2z+3z^2x+3x^2z+3y^2x+3z^2y+6xyz#

Now substitute #z=1# to get:

#(x+y+1)^3#

#=x^3+y^3+1+3x^2y+3y^2+3x+3x^2+3y^2x+3y+6xy#