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I'm trying to take a (-1) from a power to 3/2 as this:

Simplify[(-(a))^(3/2) - ((-1)^(3/2) (a)^(3/2)), a > 0]

I get as a result as expected 0. But when I try to do exactly the same with the actual functions that I need:

Simplify[(-(4 b^2 E^(2 Sqrt[b^2 - a c] h) - a c (1 + E^(2 Sqrt[b^2 - a c] h))^2))^(3/2) - ((-1)^(3/2) (4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
  a c (1 + E^(2 Sqrt[b^2 - a c] h))^2)^(3/2)), (4 b^2 E^(2 Sqrt[b^2 - a c]h) - a c (1 + E^(2 Sqrt[b^2 - a c] h))^2) > 0]

I don't get zero, I get something else... I also don't get zero using FullSimplify. I wonder if Simplify is not actually using my assumption or if there may be another reason why this is not working?

Thank you!

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  • $\begingroup$ MMA Version 8.0 gives zero to your second Simplify without problems. $\endgroup$
    – Akku14
    Commented Apr 10, 2018 at 16:35

3 Answers 3

1
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$Version

(* "11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)" *)

Use ComplexExpand which will assume that all variables are real

Simplify[(-(4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
        a c (1 + E^(2 Sqrt[b^2 - a c] h))^2))^(3/
      2) - ((-1)^(3/2) (4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
        a c (1 + E^(2 Sqrt[b^2 - a c] h))^2)^(3/2)) // 
  ComplexExpand, (4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
    a c (1 + E^(2 Sqrt[b^2 - a c] h))^2) > 0]

(* 0 *)
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You can help Mathematica as follows. Let us replace the subexpression in question by x. This is done by the rule:

rule = a c (1 + E^(2 Sqrt[b^2 - a c] h))^2 ->x + 4 b^2 E^(2 Sqrt[b^2 - a c] h)

Here is your expression:

expr = (-(4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
        a c (1 + E^(2 Sqrt[b^2 - a c] h))^2))^(3/
      2) - ((-1)^(3/2) (4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
        a c (1 + E^(2 Sqrt[b^2 - a c] h))^2)^(3/2)) /. rule

(* I (-x)^(3/2) + x^(3/2) *)

Let us now simplify and replace back:

Simplify[expr, x > 0] /.x -> a c (1 + E^(2 Sqrt[b^2 - a c] h))^2 - 
   4 b^2 E^(2 Sqrt[b^2 - a c] h)

(*  2 (-4 b^2 E^(2 Sqrt[b^2 - a c] h) + 
   a c (1 + E^(2 Sqrt[b^2 - a c] h))^2)^(3/2)   *)

Done. Have fun!

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  • $\begingroup$ Ok, so my problem is that I would like to obtain zero after using simplify, as here: $\endgroup$ Commented Apr 10, 2018 at 19:14
  • $\begingroup$ Simplify[(-(a))^(3/2) - ((-1)^(3/2) (a)^(3/2)), a > 0] $\endgroup$ Commented Apr 10, 2018 at 19:15
  • $\begingroup$ The result is zero, but not when I use the other expression. I don't understand why Simplify doesn't work in that case... $\endgroup$ Commented Apr 10, 2018 at 19:15
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FullSimplify gives the expected result:

ClearAll[a, b, c, h]
FullSimplify[(-(4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
     a c (1 + E^(2 Sqrt[b^2 - a c] h))^2))^(3/2) - 
     ((-1)^(3/2) (4 b^2 E^(2 Sqrt[b^2 - a c] h) - 
     a c (1 + E^(2 Sqrt[b^2 - a c] h))^2)^(3/2)),
   (4 b^2 E^(2 Sqrt[b^2 - a c] h) - a c (1 + E^(2 Sqrt[b^2 - a c] h))^2) > 0]

0

$Version

"11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)"

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