Factor the polynomial $x^3 − 27$ using De Moivre's theorem (Please explain solution)

Question

I was reading the book A First Course in Linear Algebra by Ken Kuttler (link to nearly identical page http://librarum.org/book/312/11) and I did not understand this part:

Q: Factor the polynomial $x^3 − 27$.

A: Solution. First find the cube roots of $27$. By the above procedure using De Moivre’s theorem, these cube roots are $3$, $3(-1/2 + i \sqrt{3}/2)$, and $3(-1/2 - i\sqrt{3}/2)$.

Unfortunately, I did look at the "above procedure" (see link if necessary) and still don't really know what happened. What exactly are the steps to get the $3$ roots? Also, thanks for the help.

Related : http://math.stackexchange.com/questions/192742/how-to-solve-x3-1 — lab bhattacharjee, Dec 16 '13 at 08:02

score 1 · Accepted Answer · answered Dec 16 '13 at 08:14

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The cube roots $z_k,\; k=0,1,2$ of $a>0$ (in your case $a=27$) are: $$z_k=\sqrt[3]{a}e^{2ik\pi/3}$$

and notice that $z_0$ is real and $z_1,z_2$ are conjugate so $$x^3-a=(x-z_0)(x-z_1)(x-z_2)=(x-z_0)(x^2-2\mathrm{Re}(z_1)x+|z_1|^2)\\=(x-\sqrt[3]{a})(x^2+\sqrt[3]{a}x+(\sqrt[3]{a})^2)$$

answered Dec 16 '13 at 08:14

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Thank you!! After some consideration (I'm not that good at math), I understood what you wrote and fully understand that part of my textbook now. :) – Yuuri Dec 16 '13 at 08:48
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You're welcome. – Dec 16 '13 at 09:01
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This needs an UV! +1 – amWhy Dec 16 '13 at 14:26

score 1 · Answer 2 · answered Dec 16 '13 at 13:51

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Note that $27=3^3$ and write $u=x/3$ to get $$ x^3 − 27=3^3(u^3-1)=3^3(u-1)(u-\omega)(u-\omega^2)=(3u-3)(3u-3\omega)(u-3\omega^2)=(x-3)(x-3\omega)(x-3\omega^2) $$

answered Dec 16 '13 at 13:51

lhf

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Factor the polynomial $x^3 − 27$ using De Moivre's theorem (Please explain solution)

2 Answers2