Example 1 In $\triangle ABC$, it is known that $x \sin A + y \sin B + z \sin C = 0$. Find the value of $(y + z \cos A)(z + x \cos B)(x + y \cos C) + (y \cos A + z)(z \cos B + x)(x \cos C + y)$.
In $\triangle A B C$, $\sin C=\sin (A+B)=\sin A \cos B+\cos A \sin B$. Substituting the known conditions, we get $x \sin A+y \sin B+z(\sin A \cos B+\cos A \sin B)=0$, which simplifies to $\sin A(x+z \cos B)=-\sin B(y+z \cos A)$.
Similarly, $\sin B(y+x \cos C)=-\sin C(z+x \cos B)$, $\sin C(z+y \cos A)=-\sin A(x+y \cos C)$,
Multiplying the three equations, we have:
$(x+z \cos B)(y+x \cos C)(z+y \cos A)=-(y+z \cos A)(z+x \cos B)(x+y \cos C)$, which simplifies to $(y+z \cos A)(z+x \cos B)(x+y \cos C)+(y \cos A+z)(z \cos B+x)(x \cos C+y)=0$.