3
$\begingroup$

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)
Improve this question
$\endgroup$
9
  • 3
    $\begingroup$ What result would you expect from it? $\endgroup$
    – MarcoB
    Commented May 25, 2018 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
    Commented May 25, 2018 at 16:06
  • $\begingroup$ Wow, so quick. I will study this answer. I think is fine. Thank you Akku14. $\endgroup$
    – Sanmuten
    Commented May 25, 2018 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
    Commented May 26, 2018 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
    Commented May 26, 2018 at 22:51

1 Answer 1

Reset to default
7
$\begingroup$

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
$\endgroup$
1
  • $\begingroup$ Surprisingly your method also works even for very large rational exponents such as 128/173 $\endgroup$
    – yarchik
    Commented May 27, 2018 at 4:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.