<s>[INST] 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.

 [/INST] $$x^3+y^3 = x^3+\color{red}{x^2y-x^2y}+y^3 $$
$$ = x^2(x+y)-y(x^2-y^2)$$
$$ = x^2(x+y)-y(x-y)(x+y)$$
$$ = (x+y)(x^2-y(x-y))$$
$$ = (x+y)(x^2-xy+y^2)$$
We can proceede similary for arbitrary odd $n$. 
</s>