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
