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.

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