2025/026) prove $a(b-c)^2 +b(c-a)^2+c(a-b)^2+8abc=$ $(a+b) (b+c) (c+a)$

 We have as on RHS (a+b) we can add 4abc to the 3rd term to get (a+b) as a factor and add 2abc to 1st and 2nd term that is distributing 8abc to get

 $a(b-c)^2 +b(c-a)^2+c(a-b)^2+8abc$ 

$=a((b-c)^2 +2bc)+b((c-a)^2+2ac)+c((a-b)^2+4ab)$

 $=a((b^2+c^2)+b((c^2+a^2))+c(a+b)^2$

 $=ab^2+ac^2+bc^2+ba^2+c(a+b)^2$

  $=ab^2+ba^2+ac^2+bc^2+c(a+b)^2$

$=ab(a+b) + c^2(a+b) + c(a+b)^2$

$=(a+b)(ab+c^2 + c(a+b)$

$=(a+b) (c^2 + ca + bc + ab)$

$= (a+b)(b+c)(c+a)$