# Binomial equations

1. Jul 21, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
My book says that one "easily" verifies that

(x+y)^n = (x + y)^(n-2)Q+(x+y)^(n-3)P where

Q = x^2 + xy +y^2

and

P = xy^2 + x^2y

2. Relevant equations

3. The attempt at a solution

I began by expanding everything into summations with binomial coefficients and it seemed like that method would work but it seemed rather far from easy.

2. Jul 21, 2007

### Hurkyl

Staff Emeritus
Try factoring.

3. Jul 21, 2007

### olgranpappy

They always say "easy" when what they mean that it can be done with a relatively small amount of work. I.e., you need no real inspiration... but it's not necessarily "easy", especially if someone tells you it's "easy." That usually just makes it "frustrating." I hate it when authors use that word. Anyways:
$$(x^2+xy+y^2)*(x+y)^{(n-2)} +(xy^2+x^2y)*(x+y)^{(n-3)}$$
rewrite the (x+y)^(n-2) in the first term as (x+y)*(x+y)^(n-3) and then factor out the (x+y)^(n-3). you get
$$(x+y)^{(n-3)}*\left[((x^2+xy+y^2)*(x+y))+xy^2+x^2y\right]$$
now it should be "easy" to show that the factor in the square bracket is just
(x+y)^3. So, we are done.