How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.


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*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$ 

*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$

*$\implies x^3 - y^3$


I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$

Let $\omega$ be a complex cube root of unity. Then $x^{3} - y^{3} = (x-y)(x- \omega y)(x-\omega^{2}y)$ since both sides vanish when $x \in \{y,\omega y,\omega^{2}y \}$ and the degrees are right. Since $1 + \omega + \omega^{2} = 0$ we have $\omega + \omega^{2} = -1.$
We also have $\omega \omega^{2} = 1$, so we have $(x - \omega y)(x-\omega^{2}y) = x^{2}+xy + y^{2}.$
