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

Question

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

My attempt at the solution:

Let $x=y=z\Rightarrow f^3 (0)=-27 \Rightarrow f(0)=-3.$

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

Let $x-y=a,y-z=b,z-x=-(a+b)$ $$\Rightarrow f(-a) \cdot f(a)+f(-b)\cdot f(b)+ f(a+b)\cdot f(-(a+b))=27.$$

I don't know what to do here :((

Thanks a real lot!

Answer 1

$f$ is continuous and cannot take the value zero, therefore it must be positive everywhere or negative everywhere. The functional equation shows that the first case is not possible.

So $f$ is negative everywhere, and we can define $g(x) = \ln(-f(x))$. Then $g$ satisfies $$ g(x-y)+g(y-z)+g(z-x) = \ln(27) $$ and you can proceed as in Find all functions $f:\mathbb{R} \to \mathbb{R},$ which is continuous in $\mathbb{R}$ then $f(x-y)+f(y-z)+f(z-x)+27=0.$.