1. Factorize $f(x, y, z)=(y+z)(z+x)(x+y)+x y z$.

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1. This is a homogeneous cyclic polynomial of degree three in $x, y, z$, and when $x=-(y+z)$, $f(x, y, z)=0$, so $f(x, y, z)$ has a factor $x+y+z$. Therefore, the other factor of $f(x, y, z)$ should be a homogeneous quadratic cyclic polynomial, hence we can set $(y+z)(z+x)(x+y)+xyz=(x+y+z)\left[k_{1}\left(x^{2}+y^{2}+z^{2}\right)+k_{2}(xy+yz+zx)\right]$, where $k_{1}, k_{2}$ are undetermined constants. Comparing the coefficients on both sides of the equation, since there are no $x^{3}, y^{3}, z^{3}$ terms on the left side, we have $k_{1}=0$. Thus, the equation simplifies to $(y+z)(z+x)(x+y)+xyz=(x+y+z) \cdot k_{2}(xy+yz+zx)$. Let $x=0, y=1, z=1$, we find $k_{2}=1$. Therefore,
$$
f(x, y, z)=(x+y+z)(xy+yz+zx).
$$