# What is the diffrence between these three commands?

Consider

F=1/((x-a)(x-b)) ;
G= 1/((x-α)*(x-β));
H=1/((x-r[1])(x-r[2])) ;


Using Apart[], the partial fraction decomposition is done only for $$F$$.

I am using version 13.1.0

What is the difference ?

sometimes you need to help Mathematica and tell it which is the variable to apply Apart on

F = 1/((x - a) (x - b))
G = 1/((x - α)*(x - β))
H = 1/((x - r[1]) (x - r[2]))


Compare

Basically, change Apart[G] to Apart[G, x] to make it work.

I am not sure if this is a feature or a bug but I've seen this before so may be it is a feature. If in doubt, always use the form Apart[expr,var]

• Thank you very much ! Being blind, there are many things that I miss Commented Jan 7 at 9:24
• @Nasser With symbolic coefficients Mathematica has no way to decide which symbols are variables and which should be understood as coefficients. Variables ordering is essential in Apart and documentation mentions that (can' t find now).
– Acus
Commented Jan 7 at 15:45