$\begin{align}  & \text{The given equation is a symmetrical expression of fourth degree} \\ & f(x)={{x}^{3}}(y-z)+{{y}^{3}}(z-x)+{{z}^{3}}(x-y) \\ & f(y)={{y}^{3}}(y-z)+{{y}^{3}}(z-y)+{{z}^{3}}(y-y) \\ & f(y)={{y}^{4}}-{{y}^{3}}z+{{y}^{3}}z-{{y}^{4}}+0 \\ & f(y)=0 \\ & \text{Therefore (}x-y)\text{ is a factor of }f(x)\text{ and hence follow } \\ & \text{that }(y-z)\text{ and (}z-x)\text{ are factors of }f(a) \\ & \text{The remaining factor will be of the }k(x+y+z) \\ & f(x)=k(x+y+z)(x-y)(y-z)(z-x) \\ & f(x)=k(x+y+z)(x-y)[yz-xy-{{z}^{2}}+xz] \\ & f(x)=k(x+y+z)\left( xyz-{{x}^{2}}y-x{{z}^{2}}+{{x}^{2}}z-{{y}^{2}}z+x{{y}^{2}}+y{{z}^{2}}-xyz \right) \\ & f(x)=k(-{{x}^{3}}y+{{x}^{3}}z-{{y}^{3}}z+x{{y}^{3}}-x{{z}^{3}}+y{{z}^{3}}) \\ & f(x)=k[{{x}^{3}}(z-y)+y(x-z)+{{z}^{3}}(y-x)] \\ & \text{So we can see }k=-1 \\ & f(x)=-(x+y+z)(x-y)(y-z)(z-x) \\\end{align}$