Example 2 Given that $f(x)$ is an $n(>0)$-degree polynomial of $x$, and for any real number $x$, it satisfies $8 f\left(x^{3}\right)-x^{6} f(2 x)-$ $2 f\left(x^{2}\right)+12=0$  
(1). Find $f(x)$.
Let the highest degree term of $f(x)$ be $a_{n} x^{n}\left(a_{n} \neq 0\right)$, then the highest degree terms of $8 f\left(x^{3}\right), x^{6} f(2 x), 2 f\left(x^{2}\right)$ are $8 a_{n} x^{3 n}, 2^{n} a_{n} x^{n+6}, 2 a_{n} x^{2 n}$, respectively. Since $2 n<3 n$, by (1) and the Polynomial Identity Theorem, we have $\left\{\begin{array}{l}3 n=n+6, \\ 8 a_{n}-2^{n} a_{n}=0 \text {. }\end{array}\right.$ Solving these, we get $n=3$, so let $f(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$. Setting $x=0$ in (1), we get $8 a_{0}-2 a_{0}+12=0$, so $a_{0}=$ -2. Substituting $f(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x-2$ into (1) and simplifying, we get $-4 a_{2} x^{8}-2 a_{1} x^{7}+\left(8 a_{2}+2-2 a_{3}\right) x^{6}$ $-2 a_{2} x^{4}+8 a_{1} x^{3}-2 a_{1} x^{2}=0$. By the Polynomial Identity Theorem, we have $\left[\begin{array}{l}-4 a_{2}=0, \\ -2 a_{1}=0, \\ 8 a_{2}+2-2 a_{3}=0 .\end{array}\right.$ Solving these, we get $a_{3}=1, a_{2}=a_{1}=$ 0, thus the required polynomial is $f(x)=x^{3}-2$.