Proving $(x+y)^n=x^n+y^n$

Question

so the problem says that if x and y are real numbers such that $xy=0$, then $(x+y)^n=x^n+y^n$. in the book there is a proof but they are using some weird constructive proof. I am to prove it using a binomial theorem without knowing the fact that $xy=0$, I realize that somehow when I break it down all my middle terms should end up $0$ while I am left with $x^n+y^n$, can anyone explain how I can show it.

Answer 1

By the binomial theorem,

$$(x + y)^n = x^n + {n \choose 1} x^{n - 1} y + {n \choose 2} x^{n - 2} y^2 + \dots + {n \choose {n - 2}} x^2 y^{n - 2} + {n \choose {n - 1}} x y^{n - 1} + y^n$$

Do you see how to extract a factor of $xy$ from every term except the first and last?

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Alternative proof: This is pretty easy by induction. If $n = 2$, then $(x + y)^2 = x^2 + 2xy + y^2 = x^2 + 2 \cdot 0 + y^2$ as desired. Then $$(x + y)^{n + 1} = (x + y)(x + y)^n = (x + y)(x^n + y^n) = x^{n + 1} + y^{n + 1} + xy(\text{things})$$

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Remark: These proofs work in an arbitrary commutative ring with $1$, which might have zero divisors. Thus this approach gives a little bit more generality than just noting that $x = 0$ or $y = 0$ (in which case, the result is quite immediate).