cubic root of negative numbers

Question

Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.

But when I checked Wolfram Alpha for $\sqrt[3]{-8}$, real it tells me it doesn't exist.

I came to trust Wolfram Alpha so I thought I'd ask you guys, to explain the sense of that to me.

Answer 1

-8 has three cube roots: $ -2 $, $1 + i \sqrt{ 3 } $ and $1 - i \sqrt{ 3 } $. So you can't answer the question "Is $ \sqrt[3]{-8} $ real" without specifying which of them you're talking about.

For some reason, WolframAlpha is only giving $1 + i \sqrt{ 3 } $ as an answer -- that looks like a bug in WolframAlpha to me.

Answer 2

When non-computers calculate the cube root of (-8), we can think of it as $(-1*8)^{1/3}$ Then we have $-1*8^{1/3} = -1*2 = -2$

Wolfram is using the polar complex form of -8 = 8cis(π) Then the cube root of this is 2cis(π/3), which is 1 + i√3 (an alternate form on Wolfram)

Incidentally, if you take $(1 + i\sqrt3)^3$, you will get -8!

Answer 3

See this. In particular, the prinicipal cube root has nonzero imaginary part.

Answer 4

Of course, you're absolutely right about $-2$ being a cubic root of $-8$.

The point might be that there are actually, three different cubic roots of $-8$, namely the roots of the polynomial $x^3+8$. One of this roots is real ($-2$), the other two are complex and conjugate of each other.

If you ask Wolfram for the cubic root of $-8$ you get one of these two non real roots, namely $1+(1.732050807568877293527446341505872366942805253810380628055...)i$.

I gather that Wolfram is instucted to choose one of the roots by some criteria that in this case leads to the exclusion of the real root. Maybe browsing the Wolfram site may help understanding what these criteria are. (My guess is that it outputs the root $\alpha=re^{i\theta}$ with smaller $\theta$ in the range $[0,2\pi)$.)