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