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Calculate the product of two radicals?

I used Basic Math Assistant to calculate the product of two radicals:

In[24]:= Power[x, (3)^-1] Power[x^2, (3)^-1]

Out[24]= x^(1/3) (x^2)^(1/3)

enter image description here

It's obvious that the result must be simplified to x. Unfortunately, Mma didn't provide such simplification...

Then, I calculated the same product as the product of two powers:

In[25]:= x^(2/3)*x^(1/3)

Out[25]= x

The result is that it must be expected.

Why the pallet application produces such an issue?
POSTED BY: Valeriu Ungureanu
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Thank you, Sander! It's interesting to observe that Mma has slightly different images for (Power) radicals and (Surd) radicals:

enter image description here
POSTED BY: Valeriu Ungureanu

Also read:

tutorial/FunctionsThatDoNotHaveUniqueValues

that you can find in the documentation.
POSTED BY: Sander Huisman

Use CubeRoot or Surd to get the real-valued roots
POSTED BY: Sander Huisman

The graph of cubic root from Wikipedia:

enter image description here

The graph of cubic root from Mathematica:

enter image description here
POSTED BY: Valeriu Ungureanu

Indeed, not a problem, a feature! Squaring and taking a cube root simply do not commute. Commutator is not equal to 0.
POSTED BY: Sander Huisman

Branch cuts are not a problem, they are a feature of mathematics.
POSTED BY: Frank Kampas

The problem appears because of the writing form of the second radical:

enter image description here
POSTED BY: Valeriu Ungureanu

Evidently, it's not a problem to obtain the correct result. But, Mma submit a strange result, especially for some users, e.g. school students.

It's interesting to see the full form of the expression:

enter image description here

In my opinion, the result have to be more rigorous, e.g. to highlight correct result for Reals and Complex numbers, separately.

POSTED BY: Valeriu Ungureanu 
In[5]:= Reduce[Power[x, (3)^-1] Power[x^2, (3)^-1] == x, x, Reals]

Out[5]= x >= 0
POSTED BY: Frank Kampas

How about :

Simplify[Power[x, (3)^-1] Power[x^2, (3)^-1], x > 0]
(* x *)
POSTED BY: Mariusz Iwaniuk

You can however use 

PowerExpand[x^(1/3) ((x^2)^(1/3))]

to simplify such kind of expressions, but be aware it might not give the right result always:

PowerExpand converts (a b)^c to a^cb^c, whatever the form of c is.

PowerExpand also converts (a^b)^c to a^(b c), whatever the form of c is.

The transformations made by PowerExpand are correct in general only if c is an integer or a and b are positive real numbers.

POSTED BY: Sander Huisman

Unfortunately this can not be simplified without getting into trouble for some numbers:

x^(1/3) ((x^2)^(1/3))
% /. x -> -1.0 - I

or, without complex numbers, just negative numbers:

x^(1/3) ((x^2)^(1/3))
% /. x -> -2.0
POSTED BY: Sander Huisman
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