assume the numbers in AP are a - d, a, a + d

Given that the sum of three numbers is 12

To find: the first three terms of AP

So,

3a = 12

a = 4

It is also given that the sum of their cube is 288

(a - d)³ + a³ + (a + d)³ = 288

a³ - d³ - 3ad(a - d) + a³ + a³ + d³ + 3ad(a + d) = 288

substituting a = 4 we get

64 - d³ - 12d(4 - d) + 64 + 64 + d³ + 12d(4 + d) = 288

192 + 24d² = 288

d = 2 or d = - 2

hence the numbers are a - d, a, a + d which is 2, 4, 6 or 6, 4, 2