> The original expression with vanilla symbols. I know Mul.flatten > doesn't combine this one, but we need to double check things if > a.base*a == a.base**(3/2) is not really valid for all x and y. But I > rather suspect that it is valid.
x**3*y*sqrt(x*sqrt(x*y)) is always (x*sqrt(x*y))**(5/2) but it is not always `(x**(3/2)*y**(1/2))**(5/2)` nor `x**(15/4)*y**(5/4)`. Neither sympy (except for factor) nor wolfram is making the mistake for vanilla symbols. -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
