14. A rectangular prism with integer edge lengths is painted red on all its surfaces, and then it is cut into small cubes with edge lengths of 1. Among them, there are 40 small cubes with two red faces, and 66 small cubes with one red face. What is the volume of this rectangular prism?
【Analysis】The small cubes with two red faces are located on the edges, and the small cubes with one red face are located at the 8 vertices. Let the number of small cubes with two red faces on the length, width, and height of the rectangular prism be $x$, $y$, and $z$ respectively. According to the problem, the sum of the number of small cubes with two red faces on the three edges meeting at the same vertex is: $x+y+z=40 \div 4=10$, (1); the sum of the number of small cubes with one red face on the three faces meeting at the same vertex is: $x y+x z+y z=66 \div 2=33$, (2); then, by combining the equations, we get: from (1) and (2), $x^{2}+y^{2}+z^{2}=10^{2}-2 \times 33=34$, (3) and $x, y, z \leqslant 5$, then we can discuss in groups by setting $x=1, 2, 3, 4, 5$.

【Solution】Let the number of small cubes with two red faces on the length, width, and height of the rectangular prism be $x$, $y$, and $z$ respectively. According to the problem,
$$
\begin{array}{l}
x+y+z=40 \div 4=10, \\
x y+x z+y z=66 \div 2=33,
\end{array}
$$

From (1) and (2), we get $x^{2}+y^{2}+z^{2}=10^{2}-2 \times 33=34$,
and $x, y, z \leqslant 5$,
From (2), we get $(x+y)(x+z)=x^{2}+33$, (4)
If $x=1, (1+y)(1+z)=1^{2}+33=2 \times 17=1 \times 34$, then $x$ and $y$ are: $2-1=1, 17-1=16$, or $1-1=0, 34-1=33$, both do not meet $x, y, z \leqslant 5$, so they are discarded;

Similarly, if $x=2, (2+y)(2+z)=2^{2}+33=1 \times 37$, then $x$ and $y$ are: $1-2=-1, 37-2=35$, which does not meet $x, y, z \leqslant 5$, so they are discarded;

If $x=3, (3+y)(3+z)=3^{2}+33=42=6 \times 7=3 \times 14=1 \times 42=21$, obviously, $3 \times 14$ and $1 \times 42$ do not meet the requirements, then $x$ and $y$ are: $6-3=3, 7-3=4$, which meets $x, y, z \leqslant 5$; if $x=4, (4+y)(4+z)=4^{2}+33=49=7 \times 7$, then $x$ and $y$ are: $7-4=3, 7-4=3$, which meets $x, y, z \leqslant 5$; and it is the same as the previous case,

If $x=5, (5+y)(5+z)=5^{2}+33=58=1 \times 58$, obviously, then $x$ and $y$ are: $1-5=-4, 58-5=54$, which does not meet $x, y, z \leqslant 5$;

Therefore, the length, width, and height of the rectangular prism are: $4+2=6, 3+2=5, 3+2=5$,
Therefore, the volume is: $6 \times 5 \times 5=150$;
Answer: The volume of this rectangular prism is 150.