4・157 $w, a, b, c$ are pairwise distinct real numbers, and it is known that there exist real numbers $x, y, z$, satisfying
$$\left\{\begin{array}{l}
x+y+z=1, \\
x a^{2}+y b^{2}+z c^{2}=w^{2}, \\
x a^{3}+y b^{3}+z c^{3}=w^{3}, \\
x a^{4}+y b^{4}+z c^{4}=w^{4} .
\end{array}\right.$$

Express $w$ in terms of $a, b, c$.
[Solution] From the first equation of the system, we know that $x, y, z$ are not all zero. Without loss of generality, assume $z \neq 0$.

From the first equation of the system, we get
$$x=1-(y+z)$$

Substituting the above equation into the last three equations of the system, we have
$$\left\{\begin{array}{l}
y\left(b^{2}-a^{2}\right)+z\left(c^{2}-a^{2}\right)=w^{2}-a^{2} \\
y\left(b^{3}-a^{3}\right)+z\left(c^{3}-a^{3}\right)=w^{3}-a^{3} \\
y\left(b^{4}-a^{4}\right)+z\left(c^{4}-a^{4}\right)=w^{4}-a^{4}
\end{array}\right.$$
(1) $\cdot\left(b^{2}+a^{2}\right)-$ (3), we get
$$\left(c^{2}-a^{2}\right)\left(b^{2}-c^{2}\right) z=\left(w^{2}-a^{2}\right)\left(b^{2}-w^{2}\right)$$
(1) $\cdot\left(b^{3}-a^{3}\right)-(2) \cdot\left(b^{2}-a^{2}\right)$, after simplification, we can obtain
$$\begin{aligned}
& (b-a)(b-c)(c-a)(a b+a c+b c) z \\
= & (b-a)(b-w)(w-a)(a b+a w+b w)
\end{aligned}$$

Since $a \neq b$, from the above equation, we know
$$\begin{aligned}
& (b-c)(c-a)(a b+a c+b c) z \\
= & (b-w)(w-a)(a b+a w+b w)
\end{aligned}$$

Since $a \neq c, b \neq c, z \neq 0, w \neq a, w \neq b$, therefore:
When $a b+a c+b c=0$, from (5), we get
$$a b+a w+b w=0,$$

which means
$$(a+b) w=-a b,$$

It is easy to see that $a+b \neq 0$ (otherwise, if $a+b=0$, from the above equation, we have $a b=0$, thus $a=0$ or $b=0$, substituting into $a+b=0$ gives $a=b=0$. This contradicts $a \neq b$). Therefore,
$$w=-\frac{a b}{a+b}$$

When $a b+a c+b c \neq 0$, (4) $\div$ (5), we have
$$\frac{(c+a)(b+c)}{a b+a c+b c}=\frac{(w+a)(b+w)}{a b+a w+b w}$$

Subtracting 1 from both sides of the above equation, we get
$$\frac{c^{2}}{a b+a c+b c}=\frac{w^{2}}{a b+a w+b w}$$

Eliminating the denominators, we have
$$(a b+a c+b c) w^{2}-(a+b) c^{2} w-a b c^{2}=0$$

which means
$$(w-c)[(a b+a c+b c) w+a b c]=0,$$

Since $w \neq c$, we get
$$(a b+a c+b c) w=-a b c$$

which means
$$w=-\frac{a b c}{a b+a c+b c} .$$

In summary, we have
$$w=\left\{\begin{array}{l}
-\frac{a b}{a+b}, \text { when } a b+a c+b c=0, \\
-\frac{a b c}{a b+a c+b c}, \text { when } a b+a c+b c \neq 0.
\end{array}\right.$$