[Example 5.2.4] Let $x, y, z, w$ be real numbers, and satisfy:
(1) $x+y+z+w=0$;
(2) $x^{7}+y^{7}+z^{7}+w^{7}=0$.

Find the value of $w(w+x)(w+y)(w+z)$.
If we take $x, y, z, w$ as the main variables, the two given conditions are not sufficient to directly determine the values of $x, y, z, w$. However, the problem only requires us to find the value of the product of four factors: $w(w+x)$, $(w+y)(w+z)$. Therefore, we still divide $x, y, z, w$ into two camps: main and auxiliary.

From condition (1), we know $x+y+z=-w$.
Thus, we consider $x, y, z$ as unknowns, i.e., the roots of a cubic equation. We set
$$
\begin{array}{c}
x y+y z+z x=p, \\
x y z=-q,
\end{array}
$$

This way, $x, y, z$ are the roots of the equation
$$
t^{3}+w t^{2}+p t+q=0
$$

From condition (2), we have $x^{7}+y^{7}+z^{7}=-w^{7}$, which inspires us to study $x^{k}+y^{k}+z^{k}$. For this, we set
$$
T_{k}=x^{k}+y^{k}+z^{k} .
$$

Then, from the given conditions, we have $T_{1}=-w, T_{7}=-w^{7}$.
$$
T_{2}=(x+y+z)^{2}-2(x y+y z+z x)=w^{2}-2 p .
$$

Since $x, y, z$ are roots of equation (1), we have
$$
x^{3}+w x^{2}+p x+q=0 \text {, }
$$

Multiplying both sides of the equation by $x^{k} \neq 0$, we get
$$
x^{k+3}=-w x^{k+2}-p x^{k+1}-q x^{k},
$$
$$
\begin{array}{l}
y^{k+3}=-w y^{k+2}-p y^{k+1}-q y^{k}, \\
z^{k+3}=-w z^{k+2}-p z^{k+1}-q z^{k},
\end{array}
$$

Similarly, we have
$$
T_{k+3}=-w T_{k+2}-p T_{k+1}-q T_{k},
$$
Adding (2), (3), and (4) gives
$$
T_{k+3}=-w T_{k+2}-p T_{k+1}-q T_{k},
$$

Noting that $T_{0}=x^{0}+y^{0}+z^{0}=3$, and $T_{1}=-w, T_{2}=w^{2}-2 p$, we get
$$
\begin{aligned}
T_{3} & =-w T_{2}-p T_{1}-q T_{0} \\
& =-w\left(w^{2}-2 p\right)-p(-w)-3 q \\
& =-w w^{3}+3 p w-3 q .
\end{aligned}
$$

Using formula (5), we can sequentially find
$$
\begin{aligned}
T_{4}= & w^{4}-4 p w^{2}+4 q w+2 p^{2}, \\
T_{5}= & -w^{5}+5 p w^{3}-5 q w^{2}-5 p^{2} w+5 p q, \\
T_{6}= & w^{6}-6 p w^{4}+6 q w^{3}+9 p^{2} w^{2}-12 p q w-2 p^{3}+3 q^{2}, \\
T_{7}= & -w^{7}+7 p w^{5}-7 q w^{4}-14 p^{2} w^{3}+21 p q w^{2}+7 p^{3} w \\
& -7 q^{2} w-7 p^{2} q .
\end{aligned}
$$

On the other hand, $T_{7}=-w^{7}$, so we have
$$
p w^{5}-q w^{4}-2 p^{2} w^{3}+3 p q w^{2}+p^{3} w-q^{2} w-p^{2} q=0 .
$$

Factoring this, we get
$$
\begin{array}{l}
(p w-q)\left(w^{4}-2 p w^{2}+q w+p^{2}\right)=0 . \\
p w-q=0, w^{4}-2 p w^{2}+q w+p^{2}=0 .
\end{array}
$$
(1) When $p w-q=0$, we calculate the product of the four factors, denoted as
$$
\begin{array}{l}
F=w(w+x)(w+y)(w+z) . \\
F=w\left[w^{3}+w^{2}(x+y+z)+w(x y+y z+z x)+x y\right] \\
=w\left[w^{3}+w^{2}(-w)+w p-q\right] \\
=w(p w-q) \\
=0 .
\end{array}
$$

Thus, $w(w+x)(w+y)(w+z)=0$.
(2) When $w^{4}-2 p w^{2}+q w+p^{2}=0$, we have $q w=2 p w^{2}-w^{4}-p^{2}$.
Noting the previous calculations,
$$
\begin{aligned}
& T_{4}=w^{4}-4 p w^{2}+4 q w+2 p^{2}, \\
T_{4}= & w^{4}-4 p w^{2}+4\left(-w^{4}+2 p w^{2}-p^{2}\right)+2 p^{2} \\
= & -3 w^{4}+4 p w^{2}-2 p^{2} \\
= & -w^{4}-2\left(w^{2}-p\right)^{2} \\
\leqslant & 0 .
\end{aligned}
$$

On the other hand, $T_{4}=x^{4}+y^{4}+z^{4} \geqslant 0$,
Thus, we must have
$$
\begin{array}{c}
-w^{4}-2\left(w^{2}-p\right)^{2}=0, \\
w^{4}+2\left(w^{2}-p\right)^{2}=0,
\end{array}
$$

Since $w, p$ are real numbers, we have
$$
w=0, w^{2}-p=0 .
$$

Thus,
$$
F=w(w+x)(w+y)(w+z)=0 .
$$

In summary, $w(w+x)(w+y)(w+z)=0$.