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$r^3+1 = (r+1)(r^2-r+1)$ 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?

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    $\begingroup$ You can do polynomial division. $\endgroup$ – Peter Dec 8 '15 at 21:56
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    $\begingroup$ Yes: $-1$ is easily seen to be a root of $r^3+1$. Then you know $r+1|r^3+1$ and you can do polynomial division to find $r^2-r+1$. $\endgroup$ – Guest Dec 8 '15 at 21:56
  • $\begingroup$ Yes, you see that the LHS has -1 as a root, and thus you divide by $(x+1)$. $\endgroup$ – fritzenbauer Dec 8 '15 at 21:57
  • $\begingroup$ It helps but is it such a mathematical proof? $\endgroup$ – diogenes Dec 8 '15 at 22:01
  • $\begingroup$ It's as much of a proof as showing that $4=2\cdot 2 $ by computing $\frac 4 2$. $\endgroup$ – YoTengoUnLCD Dec 8 '15 at 22:22
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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, $r^3 + 1 = (r+1)(r^2 - r + 1).$

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