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Evaluate $(-1)^{\frac{1}{3}}$.

I've tried to answer it by letting it be $x$ so that $x^3+1=0$. But by this way, I'll get $3$ roots, how do I get the actual answer of $(-1)^{\frac{1}{3}}$??

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What is your definition of $(\cdot)^{\frac{1}{3}}$? –  dtldarek Mar 17 '12 at 11:32
This is a related question: math.stackexchange.com/questions/109484/… –  xD13G0x Mar 17 '12 at 15:16

4 Answers 4

up vote 4 down vote accepted

It depends whether you are trying to solve it as an equation over the reals ($\mathbb{R}$) or over the complex plane ($\mathbb{C}$).
The polynomial $x^3+1$ factors as $x^3+1=(x+1)(x^2-x+1)$. So in the first case, you have exactly one solution: $x=-1$, since the second polynomial has no real roots. If you're looking for roots over $\mathbb{C}$, then you'll have three roots, since the $\sqrt[3]t$ is not a function over $\mathbb{C}$, hence all the 3 roots have equal rights to be called "the root".

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It is also ture for 1^(1/3), I think. –  eccstartup Mar 17 '12 at 13:28
@eccstartup: Sure, it's true for any $z\in\mathbb{C}$ –  Dennis Gulko Mar 17 '12 at 13:44

Have you tried looking for factors of $(x+y)^3$ ??

$(x+y)^3 = (x+y)(x^2-xy+y^2)$ therefore $(x+1)^3 = (x+1)(x^2-x+1) \Rightarrow x=-1$ is one of the root, and apply methods you learnt to find roots of quadratic equation to find other roots.

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They are all equally considered to be values of $\sqrt[3]{-1}$. There is no unique cube root, just as there is no unique square root.

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Just put it like complex numbers:

We know that $z=\sqrt[k]{m_\theta}$, so







So the answers are:




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