I want to simply $\frac{H^{\xi}\left(\frac{H}{H_0}\right)^{-\xi}}{H_0}$ this term to be $H_0^{\xi-1}$,

I run the code FullSimplify and Cancel but they do not work

Cancel[((H^\[Xi]) ((H/H0)^-\[Xi]) )/H0]
FullSimplify[((H^\[Xi]) ((H/H0)^-\[Xi]) )/H0]

How should I do that?


1 Answer 1

expr = ((H^ξ) ((H/H0)^-ξ))/H0

$\text{H0}^{\xi -1}$

  • $\begingroup$ Thank you so much! $\endgroup$
    – JieJiang
    Commented Jul 19, 2023 at 13:21

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