# Thread: Alternate expression for the cube root of 2

1. I am looking for an alternate expression for the cube root of two. The expression can include any integers, along with the constant pi (or tau). The following operations are allowed, and are the ONLY operations allowed, along with what can be derived from them:
- square roots
- trigonometric functions (like sin or cos^-1)
You are NOT allowed to use these:
- logarithms
- integrals or any calculus functions
- infinite series
- hyperbolic functions like sinh
For example, a real solution to 8x^3 - 6x - 1 = 0 can be best written in this form as cos(pi/9).
I have searched for a few days for a solution, with no luck, and a few exciting false starts that ended up quite sad. If you can provide a solution, or prove that there isn't one, that would be amazing.

2.

3. I assume you're looking for a closed form expression rather than the limit of a series. From your example, you also seem to be ruling out "a real solution to x^3 = 2."

I can tell you that just multiplication, division and square root can't reach the cube root in a finite number of steps as if you write out the powers involved, they'll always be a didactic fraction, which doesn't include 1/3. I'm 99% sure that you can include addition and subtraction in that. I'm not so sure about trig functions though.

If you want to prove it, you need to look at how various operations affect the y in 2^(x/y). You need an operation that can start with y=1 and put something other than a power of 2 there.

4. I know you can't do it with only the four arithmetic operations and square roots, otherwise, "doubling the cube" would be possible with only a straightedge and compass. It would have to rely at least partially on trigonometry.

5. Since you can somewhat recover logs and exponentials through the use of trig functions (http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Exponential_defin itions), it does seem like it might be possible. I can't really prove it either way though.

Edit: Actually, if you include the imaginary unit, then:
and
and

Not exactly simple, but it does work. Of course, if you do include the imaginary unit, there's not much reason to exclude the hyperbolic functions, since cos(i x) = cosh(x).

Can you eliminate all the imaginary numbers and do that with just integers? It doesn't really seem likely, but again, I can't prove it.

6. I do appreciate your help. There doesn't appear to be a major problem with imaginary numbers when it comes to the basic operations and square roots, but for my purposes, I don't know if I can do trigonometry with imaginary numbers. I must check out that wikipedia page and figure out what I can.

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