Question

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the largest of those three numbers?

Answer 1

Hera $3a=12$

$a=4$

Also ${(a-d)}^3+a^3+{(a+d)}^3=288,$

or $3a^3+6a{d}^2=288$

$24d^2=288-3\times 64=96$

$d^2=4$

$d=\pm 2$

Hence the numbers are $2,4,6$ or $6,4,2$