Example 1 Suppose $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=$ 0 , and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is (1990, Hefei Junior High School Mathematics Competition)
Given $a^{2}+2 a-1=0$, we know $a \neq 0$.
$$
\therefore\left(\frac{1}{a}\right)^{2}-2\left(\frac{1}{a}\right)-1=0 \text {. }
$$

From $b^{4}-2 b^{2}-1=0$, we get
$$
\left(b^{2}\right)^{2}-2\left(b^{2}\right)-1=0 \text {. }
$$

From $1-a b^{2} \neq 0$, we know $\frac{1}{a} \neq b^{2}$.
From (1) and (2), we know $\frac{1}{a}$ and $b^{2}$ are the two roots of the equation $x^{2}-2 x-1=0$, so,
$$
\frac{1}{a}+b^{2}=2, \frac{b^{2}}{a}=-1 \text {. }
$$

Therefore, $\frac{a b^{2}+b^{2}+1}{a}=\left(\frac{1}{a}+b^{2}\right)+\frac{b^{2}}{a}$
$$
=2-1=1 \text {. }
$$

Thus, the value of the original expression is 1.