The sum of three numbers in $A.P.$ is $-3$ and their product is $8$. Find those numbers.

(A) Let the three numbers in $A.P.$ be $(a-d)$, $a$, and $(a+d)$.
According to the problem, the sum is $(a-d) + a + (a+d) = -3$.
$3a = -3$, which gives $a = -1$.
The product of the numbers is $(a-d) \cdot a \cdot (a+d) = 8$.
Substituting $a = -1$, we get $(-1-d)(-1)(-1+d) = 8$.
$-1(1-d^2) = 8$, which simplifies to $d^2 - 1 = 8$.
$d^2 = 9$, so $d = \pm 3$.
If $d = 3$, the numbers are $(-1-3), -1, (-1+3)$, which are $-4, -1, 2$.
If $d = -3$, the numbers are $(-1-(-3)), -1, (-1+(-3))$, which are $2, -1, -4$.
Thus, the numbers are $-4, -1, 2$ or $2, -1, -4$.