# Taking inverse powers of an expression to that power [closed]

I'm finding a curious behavior for Mathematica 8. The code

(x^3)^(1/3)


does not want to Simplify or FullSimplify. I'd prefer not to make a rule for this, because trivial extensions such as

(x^3 y^3 z)^(1/3)


should also be able to simplify (to $xy(z)^{(1/3)}$), which is not so straightforward for a rule.

I suspect this may be related to this unanswered question.

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## closed as off-topic by Mark McClure, acl, Öskå, Michael E2, RunnyKineJul 22 '14 at 1:53

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You can see why it doesn't from this: Plot3D[ Arg[((x + I*y)^3)^(1/3)], {x, -2, 2}, {y, -2, 2} ] –  acl Jul 21 '14 at 22:27
(or Im instead of Arg, etc). Basically you need to find out what a branch cut is. –  acl Jul 21 '14 at 22:28
That's quite a pretty plot. I'm aware of branch cuts, but thanks. –  Jolyon Jul 21 '14 at 22:37
I see. Well then, the existence of branch cuts is why $(x^3)^{1/3}\neq x$, which is why Mathematica does not simplify your expression. –  acl Jul 21 '14 at 22:38

Well, it's not generally true. For example:

((-1)^3)^(1/3)
N[%]
(* Out:
(-1)^(1/3)
0.5 + 0.866025 I
*)


I suppose you could do

FullSimplify[(x^3 y^3 z)^(1/3),
Assumptions -> {x > 0, y > 0, z > 0}]
(* Out: x y z^(1/3) *)

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It appears that specifying your quantities are real in the assumptions is sufficient to pick out the real branch. Thanks. –  Jolyon Jul 21 '14 at 22:37

Others are certainly correct in pointing out that (x^3)^(1/3) is only equal to x under certain circumstances. But if you know that, and just want to simplify while assuming that x is positive, then just use PowerExpand

PowerExpand[(x^3)^(1/3)]
(*  x  *)

PowerExpand[(x^3 y^3 z)^(1/3)]

(*  x y z^(1/3)   *)

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Ah, neat! That's a useful function to know! –  Jolyon Jul 21 '14 at 22:39