I know that $i=\sqrt{-1}$. I was wondering what the $\sqrt [3] {-1}$ is.

I went on wolfram alpha, and it gave me values for $a$ and $b$ such that $\sqrt [3] {-1}=a+bi$. After some experimenting, I am almost absolutely certain we have:

$$\sqrt [3] {-1}=\frac12+\frac{\sqrt{3}}{2}i$$

I didn't know how to get to the equation above, so now that I knew the equation, I tried to work backwards. $$2\sqrt [3]{-1}=1+i\sqrt{3}$$ Square both sides:

$$4\sqrt [3]{(-1)}^2=(1+i\sqrt{3})^2$$ Now I know that $\sqrt [3]{-1}=(-1)^{1/3}$ therefore $\sqrt [3] {-1}^2=\big((-1)^{1/3}\big)^2=\big((-1)^2\big)^{1/3}=1^{1/3}=1.$

Also, using the binomial theorem (or Pascal's triangle) on the RHS, and then simplifying a little, $$4=-2+2i\sqrt{3}$$ or $3=i\sqrt{3}$ which means $i=\sqrt {3}$. Uhhh... I did something wrong, didn't I :\

Could someone please help me? I went to this question, but it didn't help, albeit similar.


Wait, the question there actually did help on second thought. I just let $\sqrt [3]{-1}=1$! which means $-1=1$. Aha!

So... now I know that step is wrong, what do we do, still?

Any help would be much appreciated. Thanks! :)

P.S. Apologies if this question is a duplicate.

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    $\begingroup$ There are three complex numbers whose cube is $-1$ $\endgroup$ – J. W. Tanner Jun 24 at 3:43
  • $\begingroup$ @J.W.Tanner Ok... seems interesting... but how do we reach those? $\endgroup$ – Feeds Jun 24 at 3:44
  • $\begingroup$ They are $-1, \frac12+\frac{\sqrt3}2i$, and its complex conjugate; are you familiar with polar coordinates for complex numbers? $\endgroup$ – J. W. Tanner Jun 24 at 3:45
  • $\begingroup$ @J.W.Tanner gulp ... no. I have heard of them, but hearing ain't enough. $\endgroup$ – Feeds Jun 24 at 3:47

The equation $x^3+1=0$ has $3$ roots. One of them is $-1$. Factoring out $x+1$ from the cubic polynomial we get $$x^3+1=(x+1)(x^2-x+1).$$ The other two cube roots of $-1$ can be found by solving the quadratic equation $$x^2-x+1=0$$ by completing the square or using the quadratic formula.

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    $\begingroup$ Oooohhhh, thank you very much! I think the other answer is also good, but this one favours best my skill level. Thanks heaps! When would you like the checkmark? :) $\endgroup$ – Feeds Jun 24 at 3:48
  • $\begingroup$ If you find an answer helpful, there's no reason to delay accepting it. $\endgroup$ – Robert Shore Jun 24 at 3:53
  • $\begingroup$ @MrPie You're welcome! Whether, which, and when to accept an answer is completely up to you. $\endgroup$ – bof Jun 24 at 3:54
  • $\begingroup$ @RobertShore There might be a reason to delay acceptance. Just because one answer is helpful, that doesn't mean an even more helpful answer won't come along. Questions with accepted answers draw less attention and are less likely to get further answers. $\endgroup$ – bof Jun 24 at 3:56
  • $\begingroup$ @RobertShore oh yeah, you're right. I'm just used to the ways of Puzzling.SE :P $\endgroup$ – Feeds Jun 24 at 4:07

Let $x=\sqrt [3] {-1}$. Then we know that $x^3=-1$. What number gives us this?

It's good to know what multiplication of complex numbers means to understand this. If you take two complex numbers, $\alpha$ and $\beta$ and multiply them, the result is: $\alpha\beta = |\alpha|\,|\beta|\,\exp\left(i( \angle\alpha + \angle \beta)\right)$. That is, we multiply the magnitudes and then sum the angles.

Ok, so now we know what we want. We know that $|x|=1$ (because, if it weren't, then $|x|^3\neq 1$). And we know that $3\angle x = \pi$. One such number is $e^{i\pi/3}$. Can you find another?

In general, for any complex number $y$, there are $n$ distinct values of $\sqrt [n] {y}$.

  • $\begingroup$ $(+1)$ This answer is a bit too advanced for me. Maybe I put my nose in places I shouldn't have. But, eh, I'm a curious person. I appreciate your efforts, nonetheless, so you can have an upvote! :) $\endgroup$ – Feeds Jun 24 at 3:49

To find all the cube roots of $-1$, you need to solve $x^3 = -1$, which can be re-written $x^3+1 = 0$.

The left hand side can be factorised into $(x+1)(x^2-x+1)$, so:

$(x+1)(x^2-x+1) = 0$

$x=-1$ or $x^2-x+1 = 0$

The quadratic can be solved using the formula to give $x = \frac{1\pm\sqrt{-3}}2 = \frac{1\pm i\sqrt{3}}2$

So the three cube roots of $-1$ are $-1, \frac{1+ i\sqrt{3}}2, \frac{1-i\sqrt{3}}2$

  • $\begingroup$ Sorry, Deepak, but @bof already beat you to it. Guess he sniped you, then, eh? Still... $(+1)$ :P $\endgroup$ – Feeds Jun 24 at 3:51
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    $\begingroup$ No worries mate. It took me a bit longer because I typed out a more complete solution. $\endgroup$ – Deepak Jun 24 at 3:51

Assuming you're not familiar with polar coordinates as suggested by @J.W.Tanner, this algebraic way might be simpler to understand.

Define a complex number to represent the cube root:

$\sqrt [3] {-1} = a + b i$


$-1 = (a + b i)^3$

which simplifies to

$-1 = a^3 - 3 a b^2 + (3a^2 b - b^3 )i$


$-1 = a^3 - 3 a b^2 $

$ 0 = 3a^2 b - b^3 $

which has solutions $a = \frac {1}{2}; b = \frac{\sqrt{3}}{2}$

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    $\begingroup$ or $a=-1$ and $b=0$ $\endgroup$ – J. W. Tanner Jun 24 at 4:05
  • $\begingroup$ or $a=\frac12; b=\frac{-\sqrt3}2$ $\endgroup$ – J. W. Tanner Jun 24 at 4:12
  • $\begingroup$ $3a^2b-b^3=0\implies3a^2=b^2$ or $b=0$; if $3a^2=b^2$ and $a^3-3ab^2=-1$ then $a^3-9a^3=-8a^3=-1$ so $a=\frac12$ $\endgroup$ – J. W. Tanner Jun 24 at 11:43

Take $-1$, whose cube is $-1$, and multiply by $\omega$ and $\omega^2$, where $\omega$ is a primitive $3$rd root of unity, to get two more cube roots of $-1$.

So, say, take $\omega=e^{\frac{2\pi i}3}$. Then $-1,-\omega, -\omega^2$ would be the three cube roots of $-1$.

This works because $\omega^3=1$.

You can read more about this here.


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