Verifying Binomial Equations Easily

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SUMMARY

The discussion focuses on verifying the binomial equation (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. The method involves expanding the equation using binomial coefficients, factoring, and simplifying terms. The author emphasizes that while the problem is labeled as "easy," it can be misleading and requires careful manipulation to arrive at the conclusion that the factor simplifies to (x+y)^3.

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Homework Statement


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


Homework Equations





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.
 
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Try factoring.
 
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:
[tex] (x^2+xy+y^2)*(x+y)^{(n-2)}<br /> +(xy^2+x^2y)*(x+y)^{(n-3)}[/tex]
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
[tex] (x+y)^{(n-3)}*\left[((x^2+xy+y^2)*(x+y))+xy^2+x^2y\right][/tex]
now it should be "easy" to show that the factor in the square bracket is just
(x+y)^3. So, we are done.
 

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