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

Let the numbers be $(a-d), a,(a+d)$.

Sum $=a-d+a+a+d=12$

$\Rightarrow 3 a=12$

$\Rightarrow a=4$

Also, $(a-d)^{3}+a^{3}+(a+d)^{3}=288$

$\Rightarrow a^{3}-d^{3}-3 a^{2} d+3 a d^{2}+a^{3}+a^{3}+d^{3}+3 a^{2} d+3 a d^{2}=288$

$\Rightarrow 3 a^{3}+6 a d^{2}=288$

$\Rightarrow 3(4)^{3}+6 \times 4 \times d^{2}=288$

$\Rightarrow 192+24 d^{2}=288$

$\Rightarrow 24 d^{2}=96$

$\Rightarrow d^{2}=4$

$\Rightarrow d=\pm 2$

When $a=4, d=2$, the numbers are $2,4,6$.

When $a=4, d=-2$, the numbers are $6,4,2$.