Let the numbers are a , a + d , a + 2d

Given that sum of three numbers =12

=> a + a + d + a + 2d = 12

=> 3(a + d) =12

=> a + d = 4

=> a = 4 - d ..................1

Again sum of their cubes is 288

=> a³ + (a + d)³ + (a + 2d)³ = 288

=> a³ + a³ +d³ +3a²d +3ad² +a³ + 8d³ + 6a²d +12ad² = 288

=> 3a³ + 9d³ +9a²d +15ad² =288

=> 3(4-d)³ + 9d³ +9(4-d)²d +15(4-d)d² = 288                {using equation 1}

=> 3(64 -d³ -48d +12d² ) + 9d³ + 9(16 + d² -8d) d  + (60 -15d)d² = 288

=> 192 - 3d³ - 144d + 36d² +9d³ + 144d + 9d³ - 72d² +60d² - 15d³ = 288

=> 24d² = 288 -192

=> 24d² =96

=> d² = 96/24

=> d² = 4

=> d = ±2 

For d = 2, a = 4 – d = 4 – 2 = 2

The numbers will be 2, 4 and 6.  

For d = - 2, a = 4 - (-2) = 4 + 2 = 6

The numbers will be 6, 4 and 2.  

Hence, the required numbers are 2, 4 and 6