The answer is no. If it is possible, then let $f(x) = g(x)h(x)$, where $g(x)$ and $h(x)$ have integral coefficients and degree less than 105. Let $\alpha_1, \alpha_2, \dots, \alpha_k$ be the complex roots of $g(x)$. For each $\alpha_j$, $(j = 1, \dots, k)$, because $g(\alpha_j)=0$, then $f(\alpha_j)=0$ which means $\alpha_j^{105}-9=0$ or $\alpha_j^{105}= 9$. It follows that the relation $\mid \alpha_j\mid = \sqrt[105]{9}$ holds for all $j=1, 2,\cdots, k$. Therefore, we have $\mid \alpha_1\alpha_2\cdots\alpha_k\mid = (\sqrt[105]{9})^k$. However, this value equals the absolute value of the constant term of $g(x)$ which is an integer by the assumption. But clearly, $(\sqrt[105]{9})^k = \sqrt[105]{3^{2k}}$ cannot be an integer when $k < 105$. This contradiction means that the assumption cannot hold.