What is the sum of the coefficients of the terms containing $x$ (Problem 105, Algebra, Gelfand) In problem 105 described below, the solution provided is:

The sum of the terms containing $x$ is the sum of the terms not
  containing y less the sum of the constant terms. There can be only one
  constant term which is the result of cubing the 1 in $ (1+x-y)^3 $. Then
  the sum of the coefficients of the terms containing $x$ (based on the
  previous problem) must be $8-1=7.$


I think his solution is wrong as it's only taking into account terms that only contain $ x $ and not also the terms that include $xy$
Is my logic correct or am I missing something?
 A: If you expand $(1 + x - y)^3$, you should obtain
$$(1 + x - y)^3 = 1 + x^3 - y^3 + 3x - 3y + 3x^2 - 3x^2y + 3y^2 + 3xy^2 - 6xy$$
The sum of the coefficients of the terms containing $x$ is $1 + 3 + 3 - 3 + 3 - 6 = 1$.  You are correct that the sum of the coefficients of the terms containing both $x$ and $y$ must be taken into account since the sum of the coefficients of those terms is $-3 + 3 - 6 = -6 \neq 0$.
A: You are right, the Problem 105 must read: "What is the sum of the coefficients of the terms containing $\color{red}{\text{only}}$ $x$?"
Note that $(1+x-y)^3$ is a trinomial and its expansion is:
$$f(x,y)=(1+x-y)^3=\sum_{i,j,k\\ i+j+k=3} {3\choose i,j,k}1^ix^j(-y)^k=\\
{3\choose 3,0,0}+{3\choose 2,1,0}x+{3\choose 2,0,1}(-y)+{3\choose 1,2,0}x^2+{3\choose 1,1,1}x(-y)+{3\choose 1,0,2}(-y)^2+{3\choose 0,3,0}x^3+{3\choose 0,2,1}x^2(-y)+{3\choose 0,1,2}x(-y)^2+{3\choose 0,0,3}(-y)^3$$
Problem 103. The sum of coefficients is:
$$f(1,1)=(1+1-1)^3=1$$
Problem 104. The sum of coefficients of the terms not containing $y$ is:
$$f(1,0)=(1+1-0)^3=8$$
Problem 105. The sum of coefficients of the terms containing $\color{red}{\text{only}}$ $x$ is:
$$f(1,0)-f(0,0)=(1+1-0)^3-(1+0-0)^3=7$$
Problem 105*. The sum of coefficients of the terms containing $x$ is:
$$f(1,1)-f(0,1)=(1+1-1)^3-(1+0-1)^3=1$$
A: For positive integer $n,$
$$(1+x-y)^n=(1+(x-y))^n=1-\sum_{r=1}^n\binom nr(x-y)^r$$
Now as the sum of coefficients of $(x-y)^r$ is $(1-1)^r=0$ for $r>0,$
So, the sum of coefficients of $$(1+(x-y))^n$$ will be $1+\sum_{r=1}^n\binom nr \cdot 0$
Similarly to exclude $x$
$(1+x-y)^n=(1-y+x)^n=(1-y)^n+$ the terms containing $x$
So, the sum of coefficients of the terms excluding $x$ will be $(1-1)^n$
Similarly the sum of coefficients of the terms excluding $y$ will be $(1+1)^n$
