Question

find three consecutive terms in an A.P whose sum is -3 and produt of their cubes is 512

Answer 1

let the three terms be a-d ,a,a+d.

sum of the three terms is -3
a-d+a+a+d=-3
3a=-3
a=-1

${( - 1 - d)}^{3} \times { - 1}^{3} \times {( - 1 + d)}^{3} = 512 \\ {(( d + 1) \times (d - 1))}^{3} = 512$
${( {d}^{2} - {1}^{2} ) }^{3} = 512 \\ {d}^{2} - 1 = 8 \\ {d}^{2} = 9$
d= -3 or +3

therefore required terms are
when d=-3
2,-1,-4

when d=3
-4,-1,2