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I have a question. It might seem silly. Suppose I have $(x^3-2)$. I know that $ 2^{1/3}$ is a root. $\frac{-1 \pm \sqrt{3}i}{2^{2/3}}$ are also roots of $(x^3-2)$.

How do you convert $\frac{-1 - \sqrt{3}i}{2^{2/3}}$ to $-(-2)^{1/3}$? Thanks.

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    "How do you convert ..." You simply do not, since $(-2)^{1/3}$ does not exist. Recall that $z^{1/3}$ is well defined when $z$ is a nonnegative real number but not for every complex number $z$ (and one can only regret that the accepted answer only reinforces this serious misconception). – 2017-02-06

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The principal value of the third root of $-2\;$ is $$(-2)^{1/3}= |-2|^{1/3}\left(\cos(\frac{2}{3}\pi) + i \sin(\frac{2}{3}\pi)\right) =\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$ Multiply with $-2$ to get $$-(-2)^{1/3}=-\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$ Now multiply your expression with $2^{1/3}$ to get $$\frac{-1 - \sqrt{3}i}{2^{2/3}}=-\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$ Therefore $$\frac{-1 - \sqrt{3}i}{2^{2/3}}=-(-2)^{1/3}$$

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    Thanks @gammatester. this helps. but how did you know the principal value of the third root of -1 ? – 2017-02-06
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    From the values of $\cos$ and $\sin$ at $\frac{1}{3}\pi.$ I have included this in the first line. Note that $\pi$ is the argument of $-1.$ – 2017-02-06
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    Why should not we believe that $(-1)^{1/3}$ is rather $\frac12(1-i\sqrt3)$, or even $-1$? For the serious problems your approach runs into, please see my comment on main. The step where you use the "identity" $$z^{1/3}w^{1/3}=(zw)^{1/3}$$ for complex, not positive real, numbers $z$ and $w$, is a clear no-no case. – 2017-02-06
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    @Did: Do you disagree, that the given value is the **principal** value of the third root of $-1$ – 2017-02-06
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    What? Please read carefully my comment (or refer to a textbook on the subject for the pitfalls of manipulating the notation $z^a$ when $a$ is not an integer). – 2017-02-06
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    @Did: See e.g. Abramowitz/Stegun 3.7.28 – 2017-02-06
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    Sure, just continue to use the "rule" $z^aw^a=(zw)^a$ for every complex numbers $(z,w)$ and soon you shall "prove" that $1=2$. Kudos. (Is A-S the textbook you refer to, to learn complex analysis? Well, if so, what can I say...) – 2017-02-06
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    @Did: No, but A/S is a handy reference. And you can immediately use the formula for $-2$ without using the value for $-1$ and your "rule". – 2017-02-06
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    Are you not using this false rule for at least one of $z$ and $w$ not positive real? Are we still in the dark about the reason why you chose this third root of $-1$ and not another one? (Sorry but as long as you do not wish to discuss the maths but simply to "defend" your post (a very curious approach, even if some on this site adopt it), I am not sure the discussion can become productive. However, I mentioned clearly why your post is out of the realm of serious, rigorous, complex analysis, and the reasons are there for anybody to read, so... everything is perfect, in a way.) – 2017-02-06
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    gammatester I would not see it as strict as @Did and say that $(-2)^{1/3}$ does not exist since it is a well-defined and unique complex number if one agrees to take a fixed branch of the complex logarithm and then for $z,\alpha \in \mathbb{C}$ sets $z^{\alpha} := \exp(\alpha \log z)$. But anyway, note that in general $\log(z_1 z_2) \neq \log z_1 + \log z_2$ and therefore the quoted "rule" is of course big blunder. – 2017-02-06
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    Using this "rule" one can "prove" that $1$ is not a real number, just replace $(-2)^{1/3}$ by $(-1)^{1/3}$ in the third line of your answer and you will get that $1=z$, where $z \in \mathbb{C}\setminus \mathbb{R}$. – 2017-02-06
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    The edit directly compute principal cube root of $-2$ without using $(-1)^{1/3}$ – 2017-02-07
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    Simply pushing the faulty step further under the rug. Sorry but *we can still see it*... – 2017-02-07