Define the polynomial $P(x)=x^{3}-36 x^{2}+51 x-97$. Starting from $P_{0}(x)=\frac{1}{81} P(9 x+1)=9 x^{3}-33 x^{2}-2 x-1$, define recursively polynomials $P_{k}(x)=$ $3 a_{k} x^{3}+3 b_{k} x^{2}+c_{k} x+d_{k}$ with integers $a_{k}, b_{k}, c_{k}, d_{k}$, where $c_{k}$ is not divisible by 3, and $P_{k+1}(x)=\frac{1}{3} P_{k}\left(3 x+\varepsilon_{k}\right)$ with $\varepsilon_{k} \in\{-1,0,1\}$ for all $k$ : Since $c_{k}$ is not divisible by 3, one can choose $\varepsilon_{k}$ such that $\varepsilon_{k} c_{k}+d_{k}$ is divisible by 3; then $\frac{1}{3} P_{k}\left(3 x+\varepsilon_{k}\right)=3 a_{k+1} x^{3}+3 b_{k+1} x^{2}+c_{k+1}+d_{k+1}$ with $a_{k+1}=9 a_{k}, b_{k+1}=9 \varepsilon_{k} a_{k}+3 b_{k}$, $c_{k+1}=3\left(3 \varepsilon_{k}^{2} a_{k}+2 \varepsilon_{k} b_{k}\right)+c_{k}, d_{k+1}=\varepsilon_{k}^{3} a_{k}+\varepsilon_{k}^{2} b_{k}+\left(\varepsilon_{k} c_{k}+d_{k}\right) / 3$, thus $c_{k+1}$ is also not divisible by 3.
If we set $x_{2004}=0$ and recursively $x_{k}=3 x_{k+1}+\varepsilon_{k}$, so $P_{k}\left(x_{k}\right)=3 P_{k}\left(x_{k+1}\right)$ for $k=2003, \ldots, 0$, then for $a=9 x_{0}+1$ the number $P(a)=81 \cdot 3^{2004} P_{2004}\left(x_{2004}\right)$ is a multiple of $3^{2008}$.