I know that you can use $\pm$ for when the answer could be either positive or negative, e.g., $x^2=16$, $x=\pm 4$.

But is there a symbol that implies that you use both the positive and the negative values? For example, I want to do something along the lines of:

$$(2/3a) \left(\sqrt[3]{2b^3 - 9abc + \sqrt{−4(b^2−3a)}} + \sqrt[3]{2b^3 - 9abc - \sqrt{−4(b^2−3a)}}\right)$$

It would be very useful to not have to write out the cube root twice and instead have a plus and minus sign before the square root.

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    $\begingroup$ I'm not sure what you mean. Do you want, for example, a way to write$$(\sqrt2-\sqrt3-\sqrt5) \times(\sqrt2-\sqrt3+\sqrt5)\times\\ (\sqrt2+\sqrt3-\sqrt5) \times(\sqrt2+\sqrt3+\sqrt5)$$concisely? $\endgroup$ – Akiva Weinberger Jan 7 '16 at 15:33
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    $\begingroup$ @Alex: You could try, $$\sum_{n=0}^1 2\big(a+(-1)^nb\big)$$ $\endgroup$ – Tito Piezas III Jan 7 '16 at 15:37
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    $\begingroup$ I 'm not sure your example exactly explain what you need, if you have a long expression why not writing $ 2 \times (a+b) + 2 \times (a- b) =2 \times (2 a)$ ? $\endgroup$ – Nizar Jan 7 '16 at 15:39
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    $\begingroup$ I don't know if it helps but my actual expression is along the lines of (2/3a) * (cube root( 2b^3 - 9abc + $\sqrt{ -4*(b^2-3a)}$ ) + cube root( 2b^3 - 9abc - $\sqrt{ -4*(b^2-3a)}$ )) and I would prefer not to write out the whole cube root each time and just write it once with a plus and minus sign before the square root sign. (sorry about the formatting) $\endgroup$ – Alex Jones Jan 7 '16 at 15:51
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    $\begingroup$ Although all answers work, I plead people to NOT abbreviate two- or three- term sums, especially with summation symbols. Conciseness at the expense of readability is not appreciated by any reader. Rather, give names to the messy expressions under the root symbols and leave the sum as it is. Of all the answers, the only one that would not make me foam at the mouth is @celtschk 's. $\endgroup$ – guest Jan 7 '16 at 22:06

Another way to write it is:

$$\frac{2}{3a}\left(R_+ + R_-\right) \text{ where } R_\pm = \sqrt[3]{2b^3-9abc\pm\sqrt{-4(b^2-3a)}}$$

  • $\begingroup$ This is probably how I would write it, too. $\endgroup$ – Akiva Weinberger Jan 7 '16 at 21:41
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    $\begingroup$ Me too. This is, in general, a really good way to show underlying similarities between structures of a formula. $\endgroup$ – Stella Biderman Jan 8 '16 at 0:06
  • $\begingroup$ Just wondering. How do you make this operable in Mathematica? $\endgroup$ – Tito Piezas III Jan 8 '16 at 16:11
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    $\begingroup$ @TitoPiezasIII: The closest I could come up with is Module[{R},R[s_]:=(2 b^3-9 a b c + s Sqrt[-4(b^2 - 3 a)])^(1/3); (2/3a)(R[+1]+R[-1])] $\endgroup$ – celtschk Jan 8 '16 at 18:25

There's no single symbol for it. So just use, $$\sum_{n=0}^1\sqrt[3]{2b^3-9abc+(-1)^n\sqrt{-4(b^2-3a)}}$$

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    $\begingroup$ You could also do $\displaystyle\sum_{\epsilon\in\{-1,1\}}\sqrt[3]{\dots+\epsilon\sqrt{\dots}}$. $\endgroup$ – Akiva Weinberger Jan 7 '16 at 21:42
  • $\begingroup$ @AkivaWeinberger how does that work? :) $\endgroup$ – Ant Jan 8 '16 at 11:07
  • $\begingroup$ @Ant $\sum_{x\in\{a,b,\dots\}}f(x)$ means $f(a)+f(b)+\dotsb$, assuming $a$, $b$, etc., are distinct (since sets don't count repeated elements). $\endgroup$ – Akiva Weinberger Jan 8 '16 at 12:57
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    $\begingroup$ @AkivaWeinberger ahah oh god of course :D I had interpreted it as $\sum_{\epsilon \in (-1,1)}$ and was wondering the meaning of that :P $\endgroup$ – Ant Jan 8 '16 at 13:04

No, and there's a good reason for it: it cannot convey the necessary information.
How would the reader know that the intention is to add?
What if you intended the "and" to be for multiplication?
You need to denote the operator somehow, and that will take care of the "and" part by itself.

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    $\begingroup$ This is the most direct and accurate answer to the core question posed. $\endgroup$ – hBy2Py Jan 8 '16 at 17:32

You can do this with multisets:

$$\sum(2×(a \pm b))$$

is the expression you want.

Basically, $\pm b$ is the multiset $\{b,-b\}$, so $a \pm b$ is $\{a+b,a-b\}$, so $2 \times (a \pm b)$ is the multiset $\{2(a+b),2(a-b)\}$, then the $\sum$ just means "take the sum."

Edit. Let me add that this notation, being not entirely standard, should never be used without prior explanation. Putting all that side, however, I want to draw attention to a cultural problem here. That problem, in short, is a general aversion to new and interesting ways of denoting our thoughts and ideas. This aversion is holding mathematics back; see here, for example. Here's how it plays out in practice.

What should happen.

The reader encounters a new and unfamiliar notation. Suddenly intrigued by the possibility of denoting his ideas more tersely and clearly, he does his best to understand the conventions of that notation and/or any deeper ideas on which it is predicated, and spends some time experimenting with it. He writes a few proofs in the new and unfamiliar language. After the new language has become sufficiently familiar, he makes a judgement regarding whether or not the benefits of the new notation outweigh the costs, and makes a conscious decision to either adopt it in his own work and writing, or not to adopt it. If he chooses the former, he is unconcerned about the potential reduction in readership, because he knows that truly elite mathematicians are intellectually flexible, and that in fact, most people are quite flexible once they've adopted the right mindset. He therefore knows that, by adopting the best possible conventions that he can, he is fundamentally doing other people a favor, and that and while this may infuriate some, nonetheless the benefits of adopting the best possible conventions outweigh the costs, and that is that.

What tends to happen.

The reader encounters a new and unfamiliar notation. The parts of his brain that are responsible for tribal thinking instantly categorize the person using the notation as an "outsider" whose opinions and ideas fundamentally don't matter (unless they're already high-up in the mathematical community, in which its an automatic movement to "what should happen"). He thinks to himself: how dare she write this kind of drivel? He starts to feel angry, and he is convinced that his anger is rational and justified. Already, whether or not the new notation could be useful to him - or to mathematics - has ceased to matter. Non-standard notation! he growls. The tools of cowards! The fact that the writer, in all likelihood, is using this notation precisely because she found it to be useful has ceased to matter. All that matters is how best to attack this new and unfamiliar experience. After composing himself for a moment, he decides on the lecturing approach. He will simply talk down to the other person, until she finally "gets" that pandering to people's inflexibility is a good and noble pursuit. He leaves a comment to the effect that: "Look, if you don't want to reach the largest possible audience, then keep on writing that way." The OP reads this comment with great sadness. They could have learned something, she thinks to herself. Instead, they have learned nothing.

Take home message.

Attitudes that justify notational or intellectual stagnation deserve to be regarded with both suspicion and sadness (in perhaps equal parts), especially when they're predicated on tribal thinking in which the ingroup is privileged and ideas from the outgroup ignored. We should be careful to reject those attitudes that, by retarding progress, make mathematics worse, while consciously adopting beliefs that make it better.

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    $\begingroup$ I personally would not understand what this notation is meant to signify (I am a professional mathematician), and I would be concerned that most readers wouldn't understand it either. I've never before seen the interpretation that $\pm$ defines a multiset. $\endgroup$ – Nate Eldredge Jan 7 '16 at 19:39
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    $\begingroup$ Downvoting because I would never expect this to be a notation about multisets, and would complain if this notation was used without an explanation that I was supposed to interpret this in this way. $\endgroup$ – Stella Biderman Jan 7 '16 at 21:43
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    $\begingroup$ Another problem is that, in set-like contexts, $\times$ usually means Cartesian product, not scalar multiplication. $\endgroup$ – David Richerby Jan 8 '16 at 0:27
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    $\begingroup$ «Every downvote I gain from here on is evidence that I am right. There really is a serious cultural problem here, and shame on those who wont admit it» is the sort of thing that raises one's crackpot index. $\endgroup$ – Mariano Suárez-Álvarez Jan 9 '16 at 5:23
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    $\begingroup$ Heh. OK. I'll add that to my CV. An enemy of progress! Do I get a superpower with that? $\endgroup$ – Mariano Suárez-Álvarez Jan 9 '16 at 6:01

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