1. Given
$$
\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}=p \text {. }
$$

Then $p^{3}+p^{2}+p=$ $\qquad$ .
Notice
$$
\begin{array}{l}
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z}, \\
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y} \\
=\frac{x+y-z}{x+y+z}, \\
p^{3}+p^{2}+p \\
=\frac{x+y-z}{x+y+z}+\frac{z+x-y}{x+y+z}+\frac{y+z-x}{x+y+z} \\
=\frac{x+y+z}{x+y+z}=1 .
\end{array}
$$