Why do you have radcan in there? Radcan does not guarantee any particular
root. Why not just rootscontract(sqrt(...)/sqrt(...))?
-s
On Mon, Aug 2, 2010 at 11:03, mok-kong shen <mok-kong.shen at t-online.de>wrote:
> rootscontract(radcan(sqrt(1-y^6)/sqrt(1-y^2)));
>