Example 3 (Identity Problem) Given the identity
$$
\begin{array}{l}
(2 x-1)^{8}-(a x+b)^{8}=\left(x^{2}+c x+d\right)^{4} \\
(a>0) .
\end{array}
$$

Find the values of the real numbers $a, b, c, d$.
Explanation: Since the equation holds for all real numbers $x$,
we can take some special values of $x$ and perform calculations.
Taking $x=0$, equation (1) becomes
$1-b^{8}=d^{4}$.
Taking $x=\frac{1}{2}$, equation (1) becomes
$$
-\left(\frac{a}{2}+b\right)^{8}=\left(\frac{1}{4}+\frac{c}{2}+d\right)^{4} \geqslant 0 .
$$

On the other hand, $-\left(\frac{a}{2}+b\right)^{8} \leqslant 0$, thus,
$$
\frac{a}{2}+b=0,
$$

which means $a=-2 b$,
and $\frac{1}{4}+\frac{c}{2}+d=0$.
Considering the coefficient of the special term $x^{8}$ in the identity, we have
$$
2^{8}-a^{8}=1 \text {. }
$$

From equation (5), we get $a^{8}=255$.
Since $a>0$, then $a=\sqrt[8]{255}$, substituting into equation (3) gives
$$
b=-\frac{\sqrt[8]{255}}{2} \text {. }
$$

At this point, the coefficients on the left side of equation (1) have been determined. Substituting them into the left side of the original equation, we get
$$
\begin{array}{l}
(2 x-1)^{8}-\left(\sqrt[8]{255} x-\frac{\sqrt[8]{255}}{2}\right)^{8} \\
=256\left(x-\frac{1}{2}\right)^{8}-255\left(x-\frac{1}{2}\right)^{8} \\
=\left(x-\frac{1}{2}\right)^{8}=\left(x^{2}-x+\frac{1}{4}\right)^{4} .
\end{array}
$$

Thus, $c=-1, d=\frac{1}{4}$, which exactly satisfy equations (4) and (2).
Another method, when $a=\sqrt[8]{255}, b=$ $-\frac{\sqrt[8]{255}}{2}$ are found, instead of substituting them into the left side of the original equation, we can substitute them into equation (2) to find $d=\frac{1}{4}$, and then substitute into equation (4) to find $c=-1$.