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:

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

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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 :\

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Could someone please help me? I went to [this question](https://math.stackexchange.com/questions/1410185/is-sqrt3-1-1), 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! :)

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