4. Given $\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}$ $=p$. Then $p^{3}+p^{2}+p=$ $\qquad$
4.1 .

From the given, we have
$$
\begin{array}{l}
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y}=\frac{x+y-z}{x+y+z}, \\
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z} .
\end{array}
$$

Then \( p^{3}+p^{2}+p=1 \).