$3 \cdot 26$ If real numbers $a, b, x, y$ satisfy $a x+b y=3, a x^{2}+b y^{2}=7, a x^{3}+$ $b y^{3}=16, a x^{4}+b y^{4}=42$, find the value of $a x^{5}+b y^{5}$.
[Solution] Given $a x^{2}+b y^{2}=7$, we have
$$
\begin{array}{l}
\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y), \\
a x^{3}+b y^{3}+x y(a x+b y)=7(x+y),
\end{array}
$$

Therefore, $16+3 x y=7(x+y)$
From $a x^{3}+b y^{3}=16$, we get $\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y)$,
$$
a x^{4}+b y^{4}+x y\left(a x^{2}+b y^{2}\right)=16(x+y),
$$

Thus, $42+7 x y=16(x+y)$
(1) $\times 7$ - (2) $\times 3$ gives $x+y=-14$, so $x y=-38$.

From
$$
\begin{array}{l}
a x^{4}+b y^{4}=42 \text {, we get }\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y), \\
a x^{5}+b y^{5}+x y\left(a x^{3}+b y^{3}\right)=42(x+y),
\end{array}
$$

Therefore,
$$
\begin{aligned}
a x^{5}+b y^{5} & =42 \times(-14)-(-38) \times 16 \\
& =20
\end{aligned}
$$