Dear Waldek, thank you for your reply. It does, however not work for this case.
f := sqrt(184726398605281*%i*sqrt(163)+14962838287027761)/(17502080*sqrt(20010)) Another related question. I there an easy way to turn (computed with type Expression(Complex(Integer)) into Expression(Integer) or even Algebraic number? If I input the above expression with sqrt(-163) instead of %i*sqrt(163), I get AlgebraicNumber as type. But I am not entering f. It's otherwise given.
That is easy to implement way of denesting, however it convers only "trivial" cases, namely when argument to root is a power.
Yes. But some people get frustrated over the simple things. I know that FriCAS has its strength somewhere else, but if I can collect some of such stuff (where I am struggeling myself) into a jfricas notebook and put it only, maybe it helps some people not to run away too quickly from FriCAS.
Extra thing: I do not know why you want to simplify roots,
Just because simple roots look nicer.
More precisely, normalize should detect "dependent roots" and what you gave is one of simplest examples of dependent roots.
I have read this while browsing the sources, but normalize did not help out of the box in my case, so I was looking for something else.
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