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So I remember scrolling through Youtube a while back when I saw this video by Blackpenredpen that was about all solutions to $$x^3=1$$Now, I will admit, I didn't think that much about it when I saw it at first, but it seems to have been bugging me because I have been thinking, "Is that even possible?" Here are my attempts at finding all solutions to$$x^3=1$$

Attempt $1$

Time: circa. $3-4$ weeks ago

We write this as$$x^3=1+0i\implies x=\sqrt[3]{1+0i}$$Now, I was thinking of using the $(a+bi)^{c+di}$ formula for this, but now that I think about it, I have no idea why I would have been doing that.

Attempt $2$

Time: circa. $4-5$ hours ago

We write $1$ as $$e^{2i\pi}=\cos(2\pi k)+i\sin(2\pi k)$$$$\implies1^{1/3}=\cos\left(\dfrac{2\pi}3\right)+i\sin\left(\dfrac{2\pi}3\right),\cos\left(\dfrac{4\pi}3\right)+i\sin\left(\dfrac{4\pi}3\right),1$$And so we have$$\cos\left(\dfrac{2\pi}3\right)\pm i\sin\left(\dfrac{2\pi}3\right)=-\dfrac12\pm\dfrac{i\sqrt2}2$$$$\therefore\sqrt[3]1=1,-\dfrac12\pm\dfrac{i\sqrt2}2$$

My question

Did I find the $3$ cube roots of $1$ correctly, or what are the $3$ cube roots of $1$?

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    $\begingroup$ Depends on your definition of cube root. If you mean "$x$ is a cube root of $y$ if and only if $x^3 = y$", well, you have a very easy test to see if you're correct, don't you? $\endgroup$
    – PrincessEev
    Commented Oct 11, 2023 at 15:01
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    $\begingroup$ The fundamental theorem of algrebra guarantees $3$ roots of $z^3-1$ and since we can easily see $\gcd(z^3-1,3z^2)=1$ , they must be distinct. The last mentioned approach to find them should be the best. An additional advice : Avoid math online videos ! $\endgroup$
    – Peter
    Commented Oct 11, 2023 at 15:04
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    $\begingroup$ Is there a reason to avoid online math videos? I agree that some are rather clickbaity and annoying, though I recall blackpenredpen being one of the good ones. $\endgroup$
    – PrincessEev
    Commented Oct 11, 2023 at 15:05
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    $\begingroup$ If you check your answers by cubing them I think you will find that $\sqrt{2}$ should be $\sqrt{3}$ . $\endgroup$
    – Ethan Bolker
    Commented Oct 11, 2023 at 15:07
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    $\begingroup$ Yet another mehtod: As obviously $x=1$ is a solution to $x^3-1=0$, perform polynomial division $(x^3-1):(x-1)$ to obtain the qudratic $x^2+x+1$. The other two cube roots of $1$ are the complex roots of this quadratic. $\endgroup$
    – Hagen von Eitzen
    Commented Oct 11, 2023 at 15:07

1 Answer 1

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  1. Yes, there are really three cube roots of $1$. This follows abstractly from the fundamental theorem of algebra. Concretely, de Moivre's formula tells you how to find all the $n^{th}$ roots of $1$ for any positive integer $n$; they are given by $\cos \frac{2 \pi k }{n} + i \sin \frac{2 \pi k}{n}, 0 \le k \le n-1$.

  2. Ethan in the comments is correct that you need a $\sqrt{3}$ in your expression. Hagen in the comments is also correct that an alternative approach is to divide $x^3 - 1$ by $x - 1$ and solve the resulting quadratic using the quadratic formula.

Once you really internalize the relationship between complex numbers and rotation the $n^{th}$ roots of $1$ are just given by rotations of a regular $n$-gon. There is a quite clear geometric picture that is maybe not emphasized as much as it should be.

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