Solution:
Let $f(x)=x^{4}+x+1$ and $g(x)=x^{4}+x^{2}-x-4$. The equation is then equivalent to
$$\frac{f(x)}{g(x)}=\frac{f(x)-x}{g(x)+x} \Longleftrightarrow x(f(x)+g(x))=0$$
Hence, $x=0$ or $f(x)+g(x)=2x^{4}+x^{2}-3=(2x^{2}+3)(x^{2}-1)=0$ which gives $-1,0,1$ as acceptable values of $x$.