# cubic root of negative numbers

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 $\sqrt{-8}$, 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.

-8 has three cube roots: $-2$, $1 + i \sqrt{ 3 }$ and $1 - i \sqrt{ 3 }$. So you can't answer the question "Is $\sqrt{-8}$ 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.

• You can get all three roots from WolframAlpha with x^3=-8 but for cbrt and ^(1/3) (like sqrt and ^(1/2)) it gives a single answer which for continuity reasons is not on the negative real line. – Henry Mar 7 '11 at 15:34
• Actually, Wolfram is giving the correct principal cube root. Mathematica is much more oriented towards continuous mathematics than discrete mathematics, which makes the extension of exponentiation to odd roots of negative numbers very out of place. – user14972 Jan 9 '15 at 16:01
• The last bit is the only blemish in an otherwise excellent answer. – J. M. isn't a mathematician Mar 11 '16 at 13:08

Although it's been two years since this question was asked, some folks might be interested to know that this behavior has been modified in WolframAlpha. If you ask for the cube root of a negative number, it returns the real valued, negative cube root. Here, I just asked for "cbrt -8", for example: Note "the principal root" button. That allows you to toggle back to the original behavior. Near the bottom, we still see information on all the complex roots. We can plot functions involving the cube root and solve equations involving the cube root and it consistently acts real valued. If you just type in an equation, it will solve it, plot both sides and highlight the intersections. Here's "cbrt(x)=sin(2x)" See this. In particular, the prinicipal cube root has nonzero imaginary part.

• I think it says the prinicipal cube root has positive imaginary part, but in practice it takes gives a non-negative real cube root of a non-negative real. In fact, looking for example at (-32)^(1/5), it takes the root with the smallest non-negative anti-clockwise rotation from the positive real axis. Looking at (-32)^(3/5) it seems to take the fifth root before cubing. – Henry Mar 7 '11 at 15:40

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 $[0,2\pi)$.)

In the years since the question was asked and answered, Wolfram introduced the $\operatorname{Surd}(x,n)$ function (Mathematica 9 circa $2012$, then Alpha) to designate the real single-valued $n^{th}$ root of $x$.

For example $\sqrt{-8}$ and $\sqrt{-243}$ result directly in $-2$ and $-3$ respectively:  The $\operatorname{Cbrt}$ function discussed in Mark McClure's answer - which had changed behavior around the same time to return the real cube root by default - appears to be identical to $\operatorname{Surd}(\,\cdot\,,3)$: 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!

Although the equation $x^3+8 = 0$ has actually three roots (real $-2$ and two conjugated complex roots) still the third root of $-8$ does not exist. Nth roots are defined only for nonnegative real values. Please, consider the $$(-8)^{1/3} = (-8)^{2/6}$$ that gives either $64^{1/6} = 2$ or $(\sqrt{ -8})^2$ = nonsense. Let $(-8)^{1/3} = (-8)^{2/6}$ you never get $(-2)$ as the result. Therefore Nth root exists only of nonnegative real numbers.