Question

find the three numbers of the AP whose sum is 24 and the sum of the cube is 1968​

Answer 1

ANSWER

LET THE THREE NUMBERS BE a,a+d and a-d

=a-d+a+a+d=24

=3a=24

=a=8

[a-d]³+[a]³+[a+d]³=1968

[a+b]³=a³+b³+3a²b+3ab²

[a-b]³=a³-b³-3a²b+3ab²

=a³+a³-d³-3a²d+3ad²+a³+b³+3a²d+3ad²=1968

3a³+2[3ad²]=1968

3a³+6ad²=1968

=3[a³+2ad²]=1968

a³+2ad²=656

=512+16d²=656

=16d²=144

d²=9

=d=±3

so the three numbers are 5,8 and 11