I am writting here because I have small math problem. I have to find all configurations of a,b,c (actual numbers) for ab(a+b)=bc(b+c)=ca(c+a) Have a nice day!
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$a b(a+b)-c a(c+a)=0$ can be factored $a (b-c) (a+b+c)=0$ similarly for the other pairs of equations $b (c-a) (a+b+c)=0$ $c(a-b)(a+b+c)=0$ so $a+b+c=0$ gives one set of solutions. There are others.
Thank You Could You explain what happened here...? ab(a+b)−ca(c+a)=0 can be factored ...? a(b−c)(a+b+c)=0
$$a b(a+b)-c a(c+a)=a(b(a+b)-c(c+a))=a\left(b a +b^2-c^2-c a\right)=a(a(b-c)+(b-c)(b+c))=a(b-c)(a+b+c)$$