2. Factorize $f(x, y, z)=(y-z)^{5}+(x-z)^{5}+(x-y)^{5}$.


2. This is a fifth-degree homogeneous cyclic polynomial in $x, y, z$, and when $y=z$, $f(x, y, z)=0$, so $f(x, y, z)$ has the factors $y-z, z-x, x-y$ (or these three factors can also be derived from $f(x, y, z)$ being an alternating polynomial). Therefore, we can set $(y-z)^{5}+(z-x)^{5}+(x-$ $y)^{5}=(y-z) \cdot(z-x) \cdot(x-y) \cdot\left[k_{1}\left(x^{2}+y^{2}+z^{2}\right)+k_{2}(x y+y z+z x)\right]$. Let $x=0, y=1, z=-1$ to get $2 k_{1}-k_{2}=15$, and let $x=2, y=1, z=0$ to get $5 k_{1}+2 k_{2}=15$, solving these gives $k_{1}=5, k_{2}=-5$, hence $f(x, y, z)=5(y-z)(z-x)(x-y) \cdot\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right)$.