Question

the sum of three numbers in AP is 12 and the sum of their cubes is 408.find the number

Answer 1

Let the terms be, a, a+d, a+2d. Given that sum is 12.

That is, a+(a+d)+(a+2d) = 12
That is, 3a+3d = 12,
Meaning, a+d=4
Since sum is 12, numbers can be (1,4,7) or (2,4,6) or (3,4,5).
But it's given that sum of cubes is 408. This is satisfied only by (1,4,7).

Answer 2

Let those numbers be a-n, a and a+n
Since sum of these numbers is 12
(a-n) + a + (a+n) =12
3a = 12
a = 4
a^3 = 64 (1)

Since, sum of their cubes is 408
(a-n)^3 + a^3 + (a+n)^3 = 408
(a-n)^3 + (a+n)^3 = 408 - 64
(a-n)^3 + (a+n)^3 = 344 (2)
But,
(a-n)^3 = a^3 - n^3 - 3an (a-n)
(a+n)^3 = a^3 + n^3 + 3an (a+n)

Putting these values and (1) in (2)
64-n^3 -3*4*n (4-n) + 64+n^3 +3*4*n (4+n) = 344
3*4*n(4+n-4+n) = 216
n(2*n)=18
n^2 = 9
n=3

All the terms are
a-n = 1
a= 4
a+n = 7