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I am doing symbolic manipulations in Mathematica where expressions like

$(-1)^{1/3}\left(-\log z \right)^{2/3}+\log z^{2/3}$

We see that clearly this expression is zero. But Mathematica does not simplifies this expression further. There are many expressions like this that involves a variety of exponent most of which are zero but I want to incorporate this automatically in the code because without evaluating it consumes a huge amount of time.

Thanks!

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    $\begingroup$ "We see that clearly this expression is zero." - not in general, no. That is, unless you're making assumptions on $z$ that you haven't mentioned. $\endgroup$
    – J. M.'s lack of A.I. ♦
    Jun 14, 2015 at 20:53
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    $\begingroup$ For example, try $z=i$ and you will find the result is not zero. $\endgroup$
    – 2012rcampion
    Jun 14, 2015 at 21:29
  • $\begingroup$ Well, the range of $z\in (0,\infty)$. probably I have to make some assumptions on z, by breaking it up between (0,1) and then from $(1,\infty)$ $\endgroup$
    – Orbifold
    Jun 14, 2015 at 21:39
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    $\begingroup$ Still not 0; maybe you haven't noticed that $(-1)^{1/3}$ is complex? Here, the principal value of the cube root is not $-1$. $\endgroup$
    – J. M.'s lack of A.I. ♦
    Jun 14, 2015 at 22:05
  • $\begingroup$ This seems related: (85893) $\endgroup$
    – Mr.Wizard
    Jun 15, 2015 at 0:39

1 Answer 1

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(-1)^(1/3) (-Log[x])^(2/3) + Log[x]^(2/3) // FullSimplify[#, x > 1] &

0

Alternatively, using the real-valued cube root of x

CubeRoot[-1] CubeRoot[(-Log[x])^2] + CubeRoot[Log[x]^2]

0

CubeRoot[-1] CubeRoot[-Log[x]]^2 + CubeRoot[Log[x]]^2

0

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