Question

The sum of three numbers in A.P. is $12$, and the sum of their
cubes is $408$; find them.

Answer 1

Let $3$ numbers in A.P. be $(a-d),a,(a+d)$
$⟹$ Sum of $3$ numbers $=12$
$a-d+a+a+d=12$
$3a=12$
$a=4$
$⟹$ sum of their cubes $=408$
${(a-d)}^3+a^3+{(a+d)}^3=408$
$a^3-d^3-3a^{2}d+3a{d}^2+a^3+a^3+d^3+3a^{2}d+3a{d}^2=408$
$3a^3+6a{d}^2=408$
$a^3+2a{d}^2=136$
${\left(4\right)}^3+2\left(4\right)d^2=136$
${8d}^2=136-64$
$8d^2=72$
$d^2=9$
$d=\pm 3$
$d=3,$ series $⟹4-3,4,4+3$
$⟹1,4,7$
$d=-3,$ series $⟹4-\left(\right),4,4-3$
$⟹7,4,1$
$\therefore 3$ numbers are $1,4$ and $7$.