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$$-4^3 = -64$$ so the third root of $-64$ should be $-4$ than. $$\sqrt[3]{-64} = -4$$ But if you calculate the third root of -64 with WolframAlpha( http://www.wolframalpha.com/input/?i=third+root+of+-64 ) you get a complex number with an imaginary part of $$3.4641016151 i$$ and a real part of $$2$$

so if the third root of $4-64$ equals $-4$ AND $2 + 3.46410162 i$ (which i know is a bit foolish) than you could actually reform it like this $$ \sqrt[3]{-64} \approx 2 + 3.46410162 i | -2$$ $$ \sqrt[3]{-64} -2 \approx -6 \approx 3.46410162 i |/3.46410162$$ $$ \frac{\sqrt[3]{-64} -2}{3.46410162} ≈ \frac{-6}{3.46410162} ≈ i$$

and this have to be totally wrong, so my question is, where exactly is the mistake?

Paul
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Darius
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    The third root is not unique, there are 3 of them. – simonzack Sep 02 '15 at 18:51
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    Even WA explains unambiguously what are the three complex third roots of $-64$. Of course, equalling two of them, namely $-4$ and $2+2\sqrt{3}i$, can only lead to chaos. – Did Sep 02 '15 at 18:52
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    In complex numbers, you can have many distinct numbers as roots of the same order. For example, $-1$ has two square roots: $i$ and $-i$. – Wojowu Sep 02 '15 at 18:53
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    There are conventions, but the one WA follows is not the one Darius follows. – Robert Israel Sep 02 '15 at 18:55
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    You can ask WA for "cube root of -64" rather than "third root of -64", and it will return $-4$. – Robert Israel Sep 02 '15 at 18:59
  • It is incorrect to say that the fact that $-4)^3=-64$ implies that $-4$ is a square root of $-64$. To get that conclusion, you would need the fact that $(-4)^3=-64$ (which is also correct) rather than $-4^3=-64$. ${}\qquad{}$ – Michael Hardy Sep 02 '15 at 19:12
  • ^ sorry, typo. I meant $-4^3=-64$. ${}\qquad{}$ – Michael Hardy Sep 02 '15 at 19:23
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    I mean, really, it's just hilarious to me that 3 people so far have downvoted. WHY?? It's an awesome question that shows the OP has really been thinking about this. It's not a "dumb" question! >:[ – layman Sep 02 '15 at 19:23
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    I don't understand the downvotes - this question is based on an error which lots of students make, and the OP has clearly thought about it. What makes this a bad question? – Noah Schweber Sep 02 '15 at 19:23
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    I really do not understand those six people who upvoted. I think the question devoted mostly to explanations of why OP does not believe that $3.646i$ equals $-2$ or something like that is off-topic here. – user2097 Sep 02 '15 at 19:31
  • @NoahSchweber I have never seen any student equalling $-4$ and $2+2\sqrt{3}i$. Have you? – Did Sep 02 '15 at 19:47
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    It's not clear at all that it's a "bad" question. The OP was having trouble understanding. A bad question is a question that shows no effort. This shows a lot of effort and shows the community that the OP has put a lot of thought into it. It's scary how many people don't think that this constitutes a good question. – layman Sep 02 '15 at 20:45
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    @MorganRodgers This site doesn't have a required level of mathematical sophistication. The OP is not claiming that $-4=2+2\sqrt{3}$ - indeed, they note that that conclusion "have to be totally wrong." What they've done is made an incorrect assumption, that polynomials have unique solutions; I think realizing that you've managed to "prove" something clearly false, and then asking where you went wrong, is perfectly appropriate. So what's the objection? – Noah Schweber Sep 02 '15 at 21:43
  • @Did that's not the error I'm talking about - the error is assuming polynomials have unique roots. (The conclusion $-4=2+2\sqrt{3}i$, by the way, is something the OP recognizes as "have to be totally wrong.") – Noah Schweber Sep 02 '15 at 21:45

4 Answers4

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If we allow complex number solutions, there are three distinct solutions to the equation $x^{3}+64=0$. One of them is $-4$. Wolfram|Alpha is simply giving you the other root(s). They are not the same, and it is incorrect to say "$-4=2+(3.46\ldots)i$", because, in $\mathbb{C}$, we define two numbers to be equal if and only if they have equal real part and equal imaginary part; we know that $2\neq-4$, so these two complex numbers have unequal real part, and so are not equal.

What this means more generally is that, although in $\mathbb{R}$ we have theorems such as "$x^{3}=y^{3} \Rightarrow x=y$", in $\mathbb{C}$ the situation is quite different: in $\mathbb{C}$, the implication does not hold in general.

Will R
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Cube roots of negative real numbers are not defined in a universally agreed-upon way. As other answers here explain, there are always three distinct complex cube roots to a nonzero number. So if you try to talk about the cube root, you have a problem.

If you know you are working with real numbers only, then since precisely one of the three (complex) cube roots of a negative real number is real, it would seem OK to declare the cube root of a negative real to be that one real cube root.

But Wolfram Alpha and other applications are going to consider complex cube roots. For them, if you ask for the cube root, they will go with the one that has the smallest angular argument. For a negative real number, this cube root ends up as a complex number in the first quadrant.

These two systems shouldn't be used simultaneously by say, choosing the first when your number is negative real, and the second when it is not real. Because then you have the problem that numbers like $\sqrt[3]{-1}$ and $\sqrt[3]{-1+0.0001i}$, which should be very close to each other if we want to respect continuity, end up being very far apart.

2'5 9'2
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Note that $x^3+64=0$ implies $(x+4)(x^2-4x+16)=0$. So you have the three cubics roots of $-64$.

Euler88 ...
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  • I have some trouble with this answer. This is like saying that the square root of $4$ is the same as the numbers that solve the equation $x^{2} - 4 = 0$, but those questions are different because they have different answers. – layman Sep 02 '15 at 20:48
  • A complex number $z$ is a $n$-th root of $a$ iff $z^n=a$. So, 2 and $-2$ are square root of 4. In the same way, -4, $2+2i\sqrt{3}$ and $2-2i\sqrt{3}$ are cubics roots of -64. The function $f(z)=\sqrt[3]{z}$ have three pre images for each $a\in \mathbb{C}$. – Euler88 ... Sep 02 '15 at 21:51
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Every complex number has three cube roots: $$ (-4)^3 = -64 $$ $$ \left( 2 \pm 2i\sqrt 3 \right)^3 = -64 $$ You can find those last two cube roots by saying $-64 = 64(\cos180^\circ+i\sin180^\circ)$, so the cube roots are $$ 4(\cos60^\circ\pm i\sin60^\circ). $$ Notice that $3\times (-60^\circ) = -180^\circ$ and for present purposes $+180^\circ$ and $-180^\circ$ are the same since they both express the same thing in polar coordinates.

The fact that $(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta) = \cos(\alpha+\beta)+i\sin(\alpha+\beta))$ perhaps deserves more attention in courses where these things are taught. In this instance in place of $\alpha+\beta$ we have $60^\circ+60^\circ+60^\circ$.

You can also use algebra, as follows. $$ x^3 = -64 $$ $$ x^3 + 64 = 0 $$ Then since we know $-4$ is one solution we have $$ (x+4)(\cdots \cdots \cdots) = 0. $$ Dividing $x^3+64$ by $x+4$ (by long division if you like), we get $$ (x+4)(x^2 - 4x + 16) = 0. $$ Therefore $$ x+4 = 0 \quad\text{or}\quad x^2 - 4x+16 = 0. $$ Then solve the quadratic equation by any of the usual methods.

Your question seems to confuse $-4^3$ with $(-4)^3$. In this case they are equal, but if you confuse $-4^2$ with $(-4)^2$ it will lead you quickly into errors.

Michael Hardy
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