10. Equation
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+ \\
\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}+\frac{x^{3}+4^{3}}{(x+4)^{3}}+\frac{x^{3}+9^{3}}{(x+9)^{3}}\right]=1
\end{array}
$$

The number of distinct non-zero integer solutions is
10.4 .

Using $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$, the original equation is
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+1+\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}-\right. \\
\left.1+\frac{x^{3}+4^{3}}{(x+4)^{3}}-1+\frac{x^{3}+9^{3}}{(x+9)^{3}}-1\right]=0 \\
\Leftrightarrow \frac{x^{3}+49 x}{(x+1)(x+4)(x+9)}- \\
\quad\left[\frac{x}{(x+1)^{2}}+\frac{4 x}{(x+4)^{2}}+\frac{9 x}{(x+9)^{2}}\right]=0 .
\end{array}
$$

Dividing both sides of the equation by $x$, and rearranging, we get
$$
x\left(x^{4}-98 x^{2}-288 x+385\right)=0 \text {. }
$$

Dividing by $x$ again, we have
$$
\left(x^{2}-31\right)^{2}-(6 x+24)^{2}=0 \text {, }
$$

which is $\left(x^{2}+6 x-7\right)\left(x^{2}-6 x-55\right)=0$.
Thus, $(x+7)(x-1)(x+5)(x-11)=0$.
Upon verification, $x_{1}=-7, x_{2}=1, x_{3}=-5, x_{4}=11$ are all roots of the original equation. Therefore, the original equation has 4 integer roots.