Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
    &(x+y)(x^2 - xy + y^2)\\
    &= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
    &= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
    &= x^3 + x^2y - x^2y - xy^2 + xy^2 + y^3\\
    &= x^3 + y^3\\
\end{split}$$
Q.E.D.
Question:
Spivak says there is an easy proof that, if I use this other theorem:
$$
x^3 - y^3 = (x-y)(x^2 + xy + y^2)
$$
then, I will also allow me to find out $x^n+y^n$ whenever $n$ is odd.
How to do this?  I fail to see how.


Maybe not easier, but given $$\frac{1-z^n}{1-z}=\sum_{i=0}^{n-1}z^i\\1-z^n=(1-z)\sum_{i=0}^{n-1}z^i\\$$ let $z=\frac{y}{x}$, then $$1-\left(\frac{y}{x}\right)^n=\left(1-\frac{y}{x}\right)\sum_{i=0}^{n-1}y^ix^{-i}$$ multiplying by $x^n$ yields $$x^n-y^n=(x-y)\sum_{i=0}^{n-1}y^ix^{n-1-i}$$
