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To any great Mathematica-matician. Why Mathematica can’t factor this

Factor[-1+x^(2/3)]

I know that Factor mainly targets polynomials but in the "Generalization" section of the help page, there is a quite similar example:

Factor[x^(2/3 s) + 2 x^s + 1]
(* (1 + x^(s/3)) (1 - x^(s/3) + 2 x^(2 s/3)) *)

I would expect it to handle my case as well and that it gives a result like this one:

(x^(1/3)-1)*(x^(1/3)+1)
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  • 3
    $\begingroup$ What result would you expect from it? $\endgroup$ – MarcoB May 25 '18 at 16:03
  • $\begingroup$ It's not quite clear what result you expect as this is not a polynomial. Factor works with polynomials. $\endgroup$ – Szabolcs May 25 '18 at 16:06
  • $\begingroup$ Wow, so quick. I will study this answer. I think is fine. Thank you Akku14. $\endgroup$ – Sanmuten May 25 '18 at 16:38
  • $\begingroup$ Why is the expected result not (x^(1/6)-1)*(x^(1/6)+1)*(x^(1/3)+1)? $\endgroup$ – Daniel Lichtblau May 26 '18 at 16:04
  • $\begingroup$ I will vote to close this because it a simple mistake. The wrong assumption you made is that your function is a polynomial in x which it is not: PolynomialQ[-1 + x^(2/3), x]. The reference page of Factor clearly says that it factors polynomials. With @Akku's answer you have a good start to dig deeper but in its current form, the question makes no sense. $\endgroup$ – halirutan May 26 '18 at 22:51
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How about that

-1 + x^(2/3) // ComplexExpand[#, TargetFunctions -> {Re, Im}] & // 
             FullSimplify // Factor // PowerExpand

(*   (-1 + x^(1/3)) (1 + x^(1/3))   *)
| improve this answer | |
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  • $\begingroup$ Surprisingly your method also works even for very large rational exponents such as 128/173 $\endgroup$ – yarchik May 27 '18 at 4:49

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