# Expression simplification

I have this term

Sqrt[(B Bc)/(A^2-B Bc)]Sqrt[B Bc+A(-A+Sqrt[A^2-B Bc])]Sqrt[-B Bc+A(A + Sqrt[A^2-B Bc])],


which actually can be simplified to B*Bc. However, it seems Mathematica cannot simplify it. Any suggestions on how I'd be able to do that? Thanks.

\$Version

"12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)"

Clear["Global*"]

expr = Sqrt[(B Bc)/(A^2 - B Bc)] Sqrt[
B Bc + A (-A + Sqrt[A^2 - B Bc])] Sqrt[-B Bc + A (A + Sqrt[A^2 - B Bc])];


Require all of the arguments of Sqrt to be positive

cons = FullSimplify[And @@ Cases[expr, Sqrt[z_] :> (z > 0), Infinity]]

(* B Bc + A Sqrt[A^2 - B Bc] > A^2 *)

Simplify[expr, cons]

(* B Bc *)

• It is better to use cons=Reduce[Cases[y, Sqrt[z_] :> (z > 0), Infinity]] yielding a > 0 && 0 < b < a^2. The FullSimplify produces a spurious wrong condition a == 0 && b > 0, which upon use in Simplify yields -b. Notice, that I replace A->a and B Bc ->b. Another way to see an issue with your method is to realize that cons produced by FullSimplify contains Sqrt the argument of which also needs to be tested for positivity. Oct 25, 2020 at 12:31

Here is why:

y=Sqrt[a (a+Sqrt[a^2-b])-b] Sqrt[a (-a+Sqrt[a^2-b])+b] Sqrt[b/(a^2-b)]
FullSimplify[y,Assumptions->a>0&&b>0&&a^2-b>0]
(*b*)

Plot[{y/.{b->-1},y/.{b->1}},{a,-2,2},
PlotTheme->{"BoldColor","Frame"},
PlotLegends->{-1,1},
PlotStyle->{Thickness[0.03],Directive[Thickness[0.01]]}]
`