Initial 293 Given the equation
$$
x^{3}-(2 a+11) x^{2}+\left(a^{2}+11 a+28\right) x-28 a=0
$$

all of whose roots are positive integers. Find the value of $a$ and the roots of the equation.
It is known that $x=a$ is a root of the original equation.
Thus, $a$ is a positive integer.
Factoring the left side of the equation, we get
$$
(x-a)\left[x^{2}-(a+11) x+28\right]=0 \text {. }
$$

Since all roots of the original equation are positive integers, then
$$
x^{2}-(a+11) x+28=0
$$

has a discriminant
$$
\Delta=(a+11)^{2}-112
$$

which should be a perfect square.
Let $(a+11)^{2}-112=k^{2}\left(k \in \mathbf{Z}_{+}\right)$. Then
$$
(a+11)^{2}-k^{2}=112 \text {, }
$$

which means $(a+11+k)(a+11-k)=112$.
Obviously, $a+11+k$ and $a+11-k$ have the same parity, and
$$
a+11+k \geqslant 11 \text {. }
$$

Since $112=56 \times 2=28 \times 4=14 \times 8$, then
$$
\left\{\begin{array}{l}
a+11+k=56,28,14, \\
a+11-k=2,4,8 .
\end{array}\right.
$$

Solving these, we get $(a, k)=(18,27),(5,12),(0,3)$.
Since $a$ is a positive integer, $a=18,5$.
When $a=18$, the roots of the original equation are $18,1,28$.
When $a=5$, the roots of the original equation are $5,2,14$.