How to prove $r^3+1 = (r+1)(r^2-r+1)$

Question

$r^3+1 = (r+1)(r^2-r+1)$ I know we can simply multiply equations in the right-hand side then we get $r^3+1$. However, is there any way to construct right-hand side without knowing it?

Answer 1

Just find the three cube roots of $-1$ as $-1, 1/2 \pm i\sqrt{3}/2$ from De Moivre's formula, and multiply:

$$r^3 = -1 \to r^3 + 1 = 0 \\ (r+1)\left(r - \frac{1}{2} - \frac{i\sqrt{3}}{2}\right)\left(r - \frac{1}{2} + \frac{i\sqrt{3}}{2}\right) = 0 \\ (r+1)\left(\left(r-\frac{1}{2}\right)^2 - \left(\frac{i\sqrt{3}}{2}\right)^2\right) = 0 \\ (r+1)\left(r^2 - \frac{2r}{2} + \frac{1}{4} + \frac{3}{4}\right) = 0 \\ (r+1)(r^2 - r + 1) = 0.$$

Hence, $r^3 + 1 = (r+1)(r^2 - r + 1).$