Example 7. Let $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=0$ and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is $\qquad$.
From the given, we have $\left(\frac{1}{a}\right)^{2}-\frac{2}{a}-1=0$, $\left(b^{2}\right)^{2}-2 b^{2}-1=0$. By the definition of roots, $\frac{1}{a}, b^{2}$ are the roots of the equation $x^{2}-2 x-1=0$. Then
$$
\begin{aligned}
\frac{1}{a}+b^{2} & =2, \frac{1}{a} \cdot b^{2}=-1 . \\
\therefore \text { the original expression } & =\left(\frac{1}{a}+b^{2}+\frac{1}{a} \cdot b^{2}\right)^{1990} \\
& =(2-1)^{1990}=1 .
\end{aligned}
$$