0

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

diogenes
  • 181
  • 2
    You can do polynomial division. – Peter Dec 08 '15 at 21:56
  • 2
    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 $r^2-r+1$. – Guest Dec 08 '15 at 21:56
  • Yes, you see that the LHS has -1 as a root, and thus you divide by $(x+1)$. – fritzenbauer Dec 08 '15 at 21:57
  • It helps but is it such a mathematical proof? – diogenes Dec 08 '15 at 22:01
  • It's as much of a proof as showing that $4=2\cdot 2 $ by computing $\frac 4 2$. – YoTengoUnLCD Dec 08 '15 at 22:22
  • As others have mentioned, you can use the Factor Theorem from college algebra (or precalculus). For examples, see my answer to Finding limit of a quotient. However, I was curious as to whether I could come up with something else. It's a bit contrived, but you can fill in all the lower powers of $r$ (both signs, so they cancel) to get $r^3 + 1 = r^3 + r^2 - r^2 + r - r + 1,$ which can be regrouped as $r^2(r+1) - r(r+1) + (r+1),$ and now you can factor out $r+1.$ – Dave L. Renfro Dec 08 '15 at 22:33

1 Answers1

2

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

John
  • 26,319