Example 3 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.
Solve the left side of the equation by factoring to get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$

Since the equation about $x$
$$
x^{2}+(a+18) x+56=0
$$

has integer roots, the discriminant $\Delta=(a+18)^{2}-224$ should be a perfect square.
Let $(a+18)^{2}-224=k^{2}\left(k \in \mathbf{N}_{+}\right)$. Then
$$
(a+18+k)(a+18-k)=224 \text {. }
$$

Obviously, $a+18+k$ and $a+18-k$ have the same parity, and $a+18+k>18$.
Since $224=112 \times 2=56 \times 4=28 \times 8$, we have
$$
\left\{\begin{array} { l } 
{ a + 1 8 + k = 1 1 2 , } \\
{ a + 1 8 - k = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
a+18+k=56, \\
a+18-k=4
\end{array}\right.\right.
$$

or $\left\{\begin{array}{l}a+18+k=28 \\ a+18-k=8 .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}a=39, \\ b=55\end{array}\right.$ or $\left\{\begin{array}{l}a=12, \\ b=26\end{array}\right.$ or $\left\{\begin{array}{l}a=0, \\ b=10\end{array}\right.$ (discard).
When $a=39$, the two roots of equation (1) are -1 and -56. At this time, the three roots of the original equation are $1, -1$, and -56.

When $a=12$, the two roots of equation (1) are -2 and -28. At this time, the three roots of the original equation are $1, -2$, and -28.