Find all functions $f:\mathbb{R} \to \mathbb{R},$ which is continuous in $\mathbb{R}$ then $f(x-y)+f(y-z)+f(z-x)+27=0.$

Question

Find all functions $f:\mathbb{R} \to \mathbb{R},$ which is continuous in $\mathbb{R}$ then $$f(x-y)+f(y-z)+f(z-x)+27=0.$$

I actually don't have any ideas to deal with it, but here is some tries:

Let $x=y,y=z,z=x$ then we have $3f(0)=-27\Rightarrow f(0)=-9.$

Let $z=y$ then $f(x-y)+f(y-x)+18=0.$

Let $y=0$ then $f(x)=-18 -f(-x).$

Also, if we let $x-y=a;y-z=b;z-x=c\Rightarrow y-x=b+c.$

Thus $f(a)+f(b+c) +18 =0; f(a)+f(b)+f(c)+27=0,$ and $a+b+c=0.$

Answer 1

First, do a translation $f(x)=g(x)-9$ then you will get $g(x-y)+g(y-z)+g(z-x)=0$. Now let $x-y=a, y-z=b$ then $z-x=-(a+b)$, putting this we get $g(a)+g(b)=-g(-(a+b))$. Now from $f(x)=-18-f(-x)$ we can show $g(x)=-g(-x)$ So we get the cauchy equation.

$g(a)+g(b)=g((a+b))$.