Question

The sum of three numbers in A.P. is $12$ and the sum of their cubes is $288$. Find the sum of their squares.

Answer 1

Let $a,a-d$ and $a+d$ is the A.P
Given $3a=12\therefore a=4$
Also ${(a-d)}^3+a^3+{(a+d)}^3=288,$
or
$3a^3+6a{d}^2=288$ or $24d^2=288-3\times 64=96$ $\therefore d^2=4$ or $d=\pm 2$
Hence the numbers are 2,4,6 or 6,4,2.