Answer:

Given,

Sum of three numbers = 12

Assume the numbers in AP are a – d, a, a + d

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

Substitute the value of a = 4,

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