Find all continuous functions with the property that f(x)f(y)f(z) = 3, for every x, y, z ∈ R, with x + y + z = 0.
I am quite new to new to functional equations and I have been stuck on this problem for quite a while.
I tried to somehow get and transform into a Cauchy equation but failed during the process.
Any help would be slightly appreciated.
First of all, note that $f(0)^3=3$, by plugging $x=y=z=0$. This way, we have that $f(0)=\sqrt[3]{3}$, so let us denote $g(x)=f(x)/\sqrt[x]{3}$. The original condition tells us that $g(x)g(y)g(z)=1$ for every $x+y+z=0$, and $g(0)=1$ by construction.
Now, by plugging $y=-x, z=0$, we have that $g(x)g(-x)g(0)=1$; or $g(-x)=g(x)^{-1}$. In particular, this tells us that $g(x)\neq 0$ for all $x$.
To conclude, note that for every $x$ and $y$, we can take $z=x+y$, so $$ g(x)g(y)g(-x-y)=1 \implies g(x)g(y)=g(x+y). $$ This is now a Cauchy functional equation. Can you conclude from here?