3. Let $x_{1}, x_{2}, \cdots, x_{6}$ be positive integers, and they satisfy the relation
$$
\begin{aligned}
x_{6} & =2288, \\
x_{n+3} & =x_{n+2}\left(x_{n+1}+2 x_{n}\right)(n=1,2,3) .
\end{aligned}
$$

Then $x_{1}+x_{2}+x_{3}=$
3. 8 .

Substituting $n=1,2,3$ into the relation
$$
\begin{array}{l}
x_{n+3}=x_{n+2}\left(x_{n+1}+2 x_{n}\right) . \\
\text { we get } x_{4}=x_{3}\left(x_{2}+2 x_{1}\right), \\
x_{5}=x_{4}\left(x_{3}+2 x_{2}\right)=x_{3}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right), \\
x_{6}=x_{3}^{2}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right)\left(x_{2}+2 x_{1}+2\right) . \\
\text { Also, } x_{6}=2288=2^{4} \times 11 \times 13, \text { and } x_{1} 、 x_{2} 、 x_{3}
\end{array}
$$

are all positive integers, so
$$
x_{3}+2 x_{2} \geqslant 3, x_{2}+2 x_{1} \geqslant 3 \text {. }
$$

Since $x_{2}+2 x_{1}+2$ is 2 more than $x_{2}+2 x_{1}$, and $2^{4} \times 11 \times 13$ only has 13 and 11 with a difference of 2 and both are factors not less than 3, thus,
$$
x_{2}+2 x_{1}+2=13 \Rightarrow x_{2}+2 x_{1}=11 \text {. }
$$

Therefore, it can only be
$$
x_{3}^{2}\left(x_{3}+2 x_{2}\right)=2^{4}, x_{3}=1 \text { or } 2 \text {. }
$$

When $x_{3}=1$, $1+2 x_{2}=2^{4}, x_{2}$ is not an integer;
When $x_{3}=2$, we get $x_{2}=1, x_{1}=5$ which meets the conditions.
Thus, $x_{1}+x_{2}+x_{3}=8$.