4. Let $x^{2}-y^{2}-z^{2}=0$. Then the factorization of $x^{3}-y^{3}-z^{3}$ into a product of linear factors is $\qquad$ .
4. $(x-y)(x-z)(2 x+y+z)$.

Let $y=a x, z=b x$, then by the given condition we get $a^{2}+b^{2}=1$.
$$
\begin{array}{l}
\therefore x^{3}-y^{3}-z^{3}=x^{3}\left(1-a^{3}-b^{3}\right) \\
=x^{3}\left(a^{2}+b^{2}-a^{3}-b^{3}\right) \\
=x^{3}(1-a)(1-b)(2+a+b) \\
=(x-a x)(x-b x)(2 x+a x+b x) \\
=(x-y)(x-z)(2 x+y+z) .
\end{array}
$$