4. Find all functions $f$, defined on the set of real numbers and taking real values, such that for any real $x, y$ and $z$ the equality $f(x y z)=f(x) f(y) f(z)-6 x y z$ holds.
Answer: $f(x)=2 x$.

Solution. Let $\mathrm{f}(1)=\mathrm{a}$. Substitute $\mathrm{y}=\mathrm{z}=1$ into the equation and we get $f(x)=a^{2} f(x)-6 x$. Therefore, $\left(a^{2}-1\right) f(x)=6 x$.

If $\mathrm{a}^{2}-1 \neq 0$, then $\mathrm{f}(\mathrm{x})=\mathrm{kx}$ for some $\mathrm{k}$. If $\mathrm{a}^{2}-1=0$, then the equation does not hold for all $\mathrm{x} \neq 0$.

We need to find for which $\mathrm{k}$ the equation holds for all $x, y, z$. kxyz=k ${ }^{3} x y z-$ $6 x y z$. When $x y z=0$, the equation always holds. When $x y z \neq 0$, we get $\mathrm{k}=\mathrm{k}^{3}-6$, or $\mathrm{k}^{3}-\mathrm{k}-6=0$. This equation has only one root 2. $k^{3}-8-k+2=0 .(k-2)\left(k^{2}+2 k+4\right)-(k-2)=0 .(k-2)\left(k^{2}+2 k+3\right)=0$. $\mathrm{k}^{2}+2 \mathrm{k}+1+2=(\mathrm{k}+1)^{2}+2>0$.

Criteria. No verification that the function fits, with a different solution method: 6 points.