<s>[INST] How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$? If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies
\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$
are in HP.
If $P,Q,R$ are in HP  $\implies Q=\frac{2PR}{P+R}.$ By this method proving that  $\frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$, if $a,b,c$ are in AP; is not straight forward.
Curiously, then we need to factorize  $(2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c)$.
The question is: How do you factorize the last expression?
 [/INST] It's tricky!
\begin{array}{l}
 2ab^2  + 2b^2 c - ac^2  - bc^2  - a^2 b - a^2 c + 2abc- 2abc =  \\ 
  \\ 
  = (2ab^2  + 2b^2 c + 2abc) - \left( {a^2 b + a^2 c + abc} \right) - \left( {ac^2  + bc^2  + abc} \right) =  \\ 
  \\ 
  = 2b(ab + bc + ac) - a(ab + bc + ac) - c(ab + bc + ac) =  \\ 
  \\ 
  =  - (ab + bc + ac)\left( {a - 2b + c} \right) \\ 
 \end{array}
</s>