Three. (25 points) Given that $a$ is a positive integer. If the roots of the equation
$$
x^{3}+(a+17) x^{2}+(38-a) x-56=0
$$

are all integers, find the value of $a$ and the integer roots of the equation.
Three, it is easy to observe that the equation has an integer root $x_{1}=1$.
The left side of the equation can be factored as
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$

Since $a$ is a positive integer, the quadratic equation in $x$
$$
x^{2}+(a+18) x+56=0
$$

has a discriminant $\Delta=(a+18)^{2}-224>0$, so it must have two distinct real roots.

Since all roots of the original equation are integers, the roots of equation (1) must also be integers. Therefore, its discriminant $\Delta=(a+18)^{2}-224$ should be a perfect square.
Let $(a+18)^{2}-224=k^{2}$ (where $k$ is a non-negative integer), then
$$
(a+18)^{2}-k^{2}=224,
$$

which means $(a+18+k)(a+18-k)=224$.
Obviously, $a+18+k$ and $a+18-k$ have the same parity, and $a+18+k \geqslant 18$.
Since $224=112 \times 2=56 \times 4=28 \times 8$, we have
$\left\{\begin{array}{l}a+18+k=112, \\ a+18-k=2\end{array}\right.$ or $\left\{\begin{array}{l}a+18+k=56, \\ a+18-k=4\end{array}\right.$ or $\left\{\begin{array}{l}a+18+k=28, \\ a+18-k=8 .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}a=39, \\ k=55\end{array}\right.$ or $\left\{\begin{array}{l}a=12, \\ k=26\end{array}\right.$ or $\left\{\begin{array}{l}a=0, \\ k=10 .\end{array}\right.$
Since $a$ is a positive integer, we have
$$
\left\{\begin{array} { l } 
{ a = 3 9 , } \\
{ k = 5 5 }
\end{array} \text { or } \left\{\begin{array}{l}
a=12, \\
k=26 .
\end{array}\right.\right.
$$

When $a=39$, equation (1) becomes $x^{2}+57 x+56=0$, whose roots are -1 and -56. In this case, the three roots of the original equation are 1, -1, and -56.

When $a=12$, equation (1) becomes $x^{2}+30 x+56=0$, whose roots are -2 and -28. In this case, the three roots of the original equation are 1, -2, and -28.