What is $\sqrt [3]{-1}$ and how does one obtain its value? 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.
EDIT:
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.
 A: 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.
A: 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 $\sqrt [n] {y}$.
A: 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:
$\sqrt [3] {-1} = a + b i$
so
$-1 = (a + b i)^3$
which simplifies to
$-1 = a^3 - 3 a b^2 + (3a^2 b - b^3 )i$
Therefore:
$-1 = a^3 - 3 a b^2 $
$ 0 = 3a^2 b - b^3 $
which has solutions $a = \frac {1}{2}; b = \frac{\sqrt{3}}{2}$
A: 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 $-1, \frac{1+ i\sqrt{3}}2, \frac{1-i\sqrt{3}}2$
A: 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.
