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

Well there are two ways that come to mind.
It is clear that when $x=y$ we have $x^3-y^3=0$. Then use long division to divide $x^3-y^3$ by $x-y$ and the result will be the equation on the right.
Another way would be to write:
$$\left(\frac{x}{y}\right)^3 - 1$$
Now we wish to find the zeros of this polynomial. These correspond to $\frac{x}{y} = 1$, $\frac{x}{y} = e^{i\frac{2\pi}{3}}$ and $\frac{x}{y} = e^{i\frac{4\pi}{3}}$.
Then we can factor the polynomial as:
$$\left(\frac{x}{y}\right)^3 - 1 = \left( \frac{x}{y} - 1 \right) \left(\frac{x}{y} - e^{i\frac{2\pi}{3}}\right) \left( \frac{x}{y} - e^{i\frac{4\pi}{3}} \right)$$
If we multiply the last two factors together we find:
$$\left(\frac{x}{y}\right)^3 - 1 = \left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} -  \frac{x}{y} \left(e^{i \frac{2\pi}{3}} + e^{i \frac{4 \pi}{3}}\right) + 1 \right)$$
$$=\left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} -  \frac{x}{y} \left( 2 \cos(2\pi/3) \right) + 1 \right) = \left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} +  \frac{x}{y} + 1 \right).$$
Thus $$\left(\frac{x}{y}\right)^3 - 1 = \left( \frac{x}{y} - 1 \right) \left(\frac{x^2}{y^2} +  \frac{x}{y} + 1 \right).$$ Multiplying by $y^3$ on both sides gives the result.
