Help with sum of coefficients please!

Question

Problem 1: "Imagine that the polynomial $(1 + x - y)^3$ is converted to the standard form. What is the sum of its coefficients?"

Problem 2 (continued): "What is the sum of the coefficients of the terms not containing $y$?"

Problem 3 (continued): "What is the sum of the coefficients of the terms containing $x$?"

I'm able to solve problem 1 & 2, but I've been stuck on 3 for hours..I'd really appreciate it if someone could explain how to solve it :D

Answer 1

One should expand $(1+x-y)^3$ as follows $$(1+x-y)^3$$$$=(1+x-y)(1+x-y)^2$$ $$=(1+x-y)(1+x^2+y^2+2x-2xy-2y)$$ $$=1+x^2+y^2+2x-2xy-2y+x(1+x^2+y^2+2x-2xy-2y)-y(1+x^2+y^2+2x-2xy-2y)$$ $$=1+x^2+y^2+2x-2xy-2y+x+x^3+xy^2+2x^2-2x^2y-2xy-y-x^2y-y^3-2xy+2xy^2+2y^2$$ $$=\color{red}{x^3-y^3-3x^2y+3xy^2+3x^2+3y^2-6xy+3x-3y+1}$$

I hope you can take it from here

Answer 2

$(1+x-y)^3=(1+x-y)(1+x-y)(1+x-y)$.

1) Choose the factor for take $x$.

2) Select the other terms for each monomial from each of the other two factors. Thus, the problem reduces to found the sum of ALL coefficients of $(1+x-y)^2$.

By simmetry, since you has 3 forms to choose the step (1), the answer is three times answer in step 2.

Answer 3

The given polynomial is $f(x,y) = (1+x-y)^3$.

• Problem 1: sum of all coefficients is $f(1,1) = 1$.
• Problem 2: sum of coefficients of the terms not containing $y$ is $f(1,0) = 2^3 = 8$.
• Problem 3: sum of coefficients of the terms containing $x$ is $f(1,1) - f(0,1) = 1 - 0 = 1$.