excuse my absence of knowledge and expertise in sdrta.net,but come me it would came normally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.

You are watching: Can you cube root a negative number

But as soon as I confirm Wolfram Alpha for $sqrt<3>-8$, genuine it speak me that doesn"t exist.

I pertained to trust Wolfram Alpha for this reason I thought I"d ask girlfriend guys, to define the feeling of the to me.



-8 has actually three cube roots: $ -2 $, $1 + ns sqrt 3 $ and $1 - i sqrt 3 $. So you can"t answer the inquiry "Is $ sqrt<3>-8 $ real" there is no specifying which of castle you"re talking about.

For some reason, WolframAlpha is only providing $1 + i sqrt 3 $ as solution -- the looks choose a an insect in WolframAlpha to me.


$egingroup$ Actually, Wolfram is providing the correct primary cube root. sdrta.netematica is much more oriented towards continuous sdrta.netematics than discrete sdrta.netematics, which makes the expansion of exponentiation to odd root of negative numbers really out that place. $endgroup$
Although it"s been two years due to the fact that this question was asked, part folks might be interested to know that this actions has to be modified in WolframAlpha. If friend ask for the cube root of a negative number, it return the actual valued, an unfavorable cube root. Here, I just asked for "cbrt -8", for example:


Note "the primary root" button. That permits you come toggle earlier to the original behavior. Near the bottom, us still see information on all the facility roots.


We have the right to plot features involving the cube root and also solve equations entailing the cube root and it repetitively acts genuine valued. If friend just type in one equation, it will fix it, plot both sides and also highlight the intersections. Here"s "cbrt(x)=sin(2x)"




Of course, you"re absolutely right around $-2$ gift a cubic source of $-8$.

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

I gather the Wolfram is instucted to pick one the the root by some criteria the in this instance leads to the exemption of the real root. Possibly browsing the Wolfram website may help understanding what these criteria are. (My guess is that it outputs the source $alpha=re^i heta$ with smaller $ heta$ in the selection $<0,2pi)$.)

In the years because the concern was asked and answered, Wolfram presented the $operatornameSurd(x,n)$ role (sdrta.netematica 9 circa $2012$, climate Alpha) come designate the genuine single-valued $n^th$ source of $x$.

For example $sqrt<3>-8$ and also $sqrt<5>-243$ an outcome directly in $-2$ and also $-3$ respectively:



The $operatornameCbrt$ duty discussed in note McClure"s answer - i m sorry had changed behavior about the very same time to return the real cube source by default - shows up to be similar to $operatornameSurd(,cdot,,3)$:


When non-computers calculation the cube root of (-8), we have the right to think that it as $(-1*8)^1/3$Then we have actually $-1*8^1/3 = -1*2 = -2$

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

Incidentally, if you take $(1 + isqrt3)^3$, you will get -8!

Although the equation $x^3+8 = 0$ has actually actually 3 roots (real $-2$ and also two conjugated complex roots) tho the third root of $-8$ does no exist. Nth root are defined only for nonnegative real values. Please, think about the $$(-8)^1/3 = (-8)^2/6$$ that gives either $64^1/6 = 2$ or $(sqrt -8)^2 $ = nonsense. Allow $(-8)^1/3 = (-8)^2/6$ girlfriend never acquire $(-2)$ as the result. As such Nth root exists only of nonnegative real numbers.

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