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.

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 $-1, \frac{1+ i\sqrt{3}}2, \frac{1-i\sqrt{3}}2$
