What is the principal cubic root of $-8$?

According to my book it should be a real number, and according to WolframAlpha it should be $1+1.73i$

• Commented May 31, 2015 at 14:19
• Commented Jun 1, 2015 at 21:39

The cubic roots of $-8$ are: $$2e^{i\pi/3}=2(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}) \qquad 2e^{i\pi}=-2 \qquad 2e^{i2\pi/3}=2(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2})$$

it seems that WolframAlpha consider as principal root the root with the minimum value of the argument, i.e $e^{i\pi/3}$, but usually, if there is a real root this is considered the principal root.

• This is a good explanation only a detail: W|A considers as principal root the one obtained from the principal argument (which W|A seems to take in $(-\pi, \pi]$) so minimum value of the argument is potentially misleading.
– quid
Commented May 31, 2015 at 13:29
• I'm not sure that WA take values in $(-\pi,\pi]$, but, in this case it seems that WA defines as principal value the value with the minimum positive value of the argument. Commented May 31, 2015 at 14:00
• I am pretty sure though, as I tested it. Also see GEdgar's answer. If you want to see for yourself let it compute the cube root of $-i$, it will not give $i$ as principal value.
– quid
Commented May 31, 2015 at 14:19

Mathematicians (as opposed to elementary textbook writers) consider complex numbers. To make the principal value of the cube root continuous in as large a region of the complex plane as possible, and to have it agree with the conventional cube root for positive numbers, they choose the cube root with argument in $(-\pi/3 , \pi/3]$. This fits together with a systematic choice of principal value for logarithm, arctangent, and others.

There are three complex cubic roots of $-8$.

Namely $2 e^{i \pi/3}$, $2 e^{i (\pi+2 \pi) /3}$, $2 e^{i (\pi+4 \pi) /3}$. You get this using the polar form(s) of $-8$, that is $8 e^{i \pi}$ and more generally $8 e^{i (\pi+ 2k\pi)}$ for $k$ and integer.

The middle one is $-2$ and the first one is what Wolfram Alpha gives.

The rational for W|A making that choice is that it considers as principal root the one where the principal argument of $-8$, that is $\pi$, is divided by $3$.

The rational of your book should be that they want to be cubic roots of reals to be real.

It is a matter of definition/convention. In some sense I prefer the former, but would use the latter in the right context too.

In $\mathbb R$, the cube root of $-8$ is $-2$. There is no need for the concept of a "principal root".

So the term "principal cube root" implies that the field of interest is $\mathbb C$, not $\mathbb R$. And then, according to this Wikipedia article, the principal cube root of $z$ is usually defined as $\exp(\frac13\log z)$, where the principal branch of the logarithm is taken. This interpretation gives $1 + i\sqrt 3$ as the principal cube root of $-8$.

By the way, did you notice that Wolfram Alpha has this covered? It lets you choose:

Assuming the principal root | Use the real‐valued root instead

EDIT 1:

I was trying to say about the real part of the complex root. We need be totally lost when helped by the graph of the function. The real root still hangs around an expected location.

If you sketch a graph of $$y = x^3 + 8$$ , it is seen cutting x-axis not only at real x = -2 but the complex roots have real parts coming closest/ farthest to x-axis near about x= 1 when not actually cutting x-axis... depending on lack of harmonicity.

At these points you have complex roots $$1 \pm i \sqrt {3}$$ exactly. \$ A view in Complex plane shows all the three positions of vector tip in a WA plot ( not shown here).

To illustrate, I took an irregular graph the red lines point to location of real part of existence of the complex roots.

There are 3 ways non-contacting and intersecting/.. or .. convex/concave $$\cup, \cap$$ sort of shaped curves aspected to x-axis in a graph appear e.g., like when $$y=\cosh x, x (1-x), x^6$$ respectively. Correspondingly imaginary solutions are odd multiples of $$i \pi/2$$ along with the regular real roots$$(0,1)$$ appear..the former non touching cup shapes produce complex roots. The tangential roots are well known.

For $$y(x)= x^3+8 , y^{''}=6x=0$$ a maximum and minimum of y=(x) should occur to the right and left of $$x=0$$ and complex roots whose real parts should appear nearer to the red abcissas. Exact complex roots can be calculated numerically accurately of course by Newton-Raphson like procedures however.

• I don't understand what you mean by y = x^3 + 8 "comes closest to x-axis near about x = 1". Huh?? I do understand why that would be nice, seeing how 1 is the real part of the two nonreal zeros. But y(1) = 9 "seems close to x-axis compared to y at other x values"? What am I missing here? Commented Dec 12, 2021 at 7:36
• Sorry. The real root presents no problem but about complex root I should have explained better at that time. Re-edited the answer. Please feel free to further comment further if lacking in clarity ( which perhaps it still does.) Commented Dec 12, 2021 at 20:41

Consider $$x^3 = - 8$$. It "seems" we can just say $$x = -2$$ because $$(- 2)^3 = -8$$ but $$- 2$$ is not the official Primary Solution!

Using De Moivre's theorem again: $$x^3 = - 8$$ $$( r \operatorname{cis}(\theta) )^3 = -8$$ $$r^3 \operatorname{cis}(3\theta) = 8\operatorname{cis}( 180^\circ + 360^\circ n)$$ $$r^3 = 8 \text{ and } 3\theta = 180^\circ + 360^\circ n$$ $$r = 2 \text{ and } \theta = 60^\circ + 120^\circ n = 60^\circ, 180^\circ, 300^\circ$$

$$x_1 = 2\operatorname{cis}( 60^\circ ) = 1 + i\sqrt3$$ $$x_2 = 2\operatorname{cis}(180^\circ) = - 2$$ $$x_3 = 2\operatorname{cis}(240^\circ) = 1 - i\sqrt3$$

The Primary Solution is $$x_1 = 1 + i\sqrt3 ≈ 1 + 1.732i$$ So the cube root of $$-8$$ or $$(-8)^{\frac13}=1+1.732i$$ and NOT $$-2$$.

• Welcome to MSE. This answer would improve substantially by using MathJax to typset. It's very easy to use; a guide can be found here math.meta.stackexchange.com/questions/5020/… Commented Aug 22, 2020 at 23:46