# How do you find the cube roots of just a number?

I was told to write the $3$ cube roots of $27$ in rectangular form how is this to be done? I know how to do cube roots on something like $5+0i$ in polar form, but how do you do it with just a number?

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If you know how to find the cube roots of $5+0i$ in polar form, you're most of the way there already. $27=27+0i$, so you can use the same techniques on it as on $5+0i$ to get the polar form of the cube roots. To get to rectangular form from polar form, evaluate the polar form—that is, $$re^{i\theta}=r(\cos\theta+i\sin\theta)=r\cos\theta+(r\sin\theta)i.$$

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$z^\frac{1}{n} = +\sqrt[n]{\vert z\vert} e^{ i \frac{\arg (z) + 2k\pi }{n} }$, where $\arg(z)$ is the principal argument of the complex number $z$, and $k = {0, 1, ... n-1}$. In this case, since the complex number $z$ is purely real, $\arg(z) = 0$. Therefore we are left with:

$27^\frac{1}{3} = \sqrt[3]{27} e^{i\frac{2k\pi}{3} }$, which, iterating over $k$, gives us the three roots $3, 3e^{i\frac{2\pi}{3}}$, and $3e^{i\frac{4\pi}{3}}$. Now, to express them in rectangular form, simply use Euler's formula; $e^{ix} = \cos(x) + i \sin(x)$.

Your final answers should then be $3, \frac{3}{2}(-1+i\sqrt{3})$, and $\frac{3}{2}(-1-i\sqrt{3})$.

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@Cirvante. By the way, maybe you should take a look at meta.math.stackexchange.com/questions/106/… . – a.r. Sep 26 '10 at 9:40
@Agusti Roig Don't worry, I really don't care a ton for the actual answer I need to find out how to do it so that I can do any different problem on the test. – chromedude Sep 27 '10 at 10:57

Yes, a number has three cube roots. Think about how you would factor the polynomial $x^3-27$.

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yes, of course. – jericson Sep 26 '10 at 3:50