The sum of three numbers in $A.P.$ is $12$ and the sum of their cubes is $288$. Find those numbers.

(A) Let the three numbers in $A.P.$ be $(a-d)$,$a$,and $(a+d)$.
According to the problem,their sum is $(a-d) + a + (a+d) = 12$.
This simplifies to $3a = 12$,so $a = 4$.
The numbers are $(4-d)$,$4$,and $(4+d)$.
The sum of their cubes is $(4-d)^3 + 4^3 + (4+d)^3 = 288$.
Expanding the cubes: $(64 - 48d + 12d^2 - d^3) + 64 + (64 + 48d + 12d^2 + d^3) = 288$.
Combining terms: $192 + 24d^2 = 288$.
$24d^2 = 96$,which gives $d^2 = 4$,so $d = \pm 2$.
If $d = 2$,the numbers are $(4-2), 4, (4+2)$,which are $2, 4, 6$.
If $d = -2$,the numbers are $(4-(-2)), 4, (4+(-2))$,which are $6, 4, 2$.
Thus,the numbers are $2, 4, 6$ or $6, 4, 2$.