CAS and exponents issues

[parisse]

(09-28-2018 06:48 AM)sasa Wrote:  If I do not mistake, you are mathematician Bernard Parisse, who have made several revisions of CAS for HP, including current for the Prime. If not, please sorry for identity confusion.

I have reported concrete expressions where Prime failed. This one aims to
\(2^{1/3} \cdot 3^{1/3}=2^{1/3} \cdot 3^{1/3}\), instead simply \(6^{1/3}\)

This is elementary example were CAS in Prime fails...

Why do you say it fails? The CAS is not using your prefered representation but there is nothing wrong.
Please note that simplify(2^(1/3)*3^(1/3)-6^(1/3)) returns 0, i.e. you will not accidentally divide by an expression that does not look like 0 but is 0, and that is what really matters when you run computations. If you simplify(2^(1/3)*3^(1/3)), it remains as is on the Prime because HP does not want to see rootofs, but in Xcas it will be rewritten as a rootof, i.e. as an element of an algebraic extension of Q that contains 2^(1/3) and 3^(1/3), that's the best way I know to make sure you don't miss 0 simplifications.

Quote:As I'm one who made lite version of CAS three decades ago from ground for my own pleasure, I'm hardly novice. Nowadays CAS used in calculators and desktop computers are quite complex that even slightly change in code, step orders and settings in general may bring such inconsistence even with elementary expressions...

Can we test your lite version of CAS somewhere?
It is probably not implemented with the same philosophy as Xcas. Xcas was primarily implemented to run big computations that a human would never make by hand (i.e. mostly big integers and polynomials computations).