Question

please give its full solution

Answer 1

Let the three numbers be a-d,a and a + d.

1st condition:

Given that sum of three numbers = 12

a - d + a + a + d = 12

3a = 12

a = 4.


2nd condition:

Given sum of their cubes = 288.

(a - d)^3 + a^3 + (a + d)^3 = 288

a^3 - d^3 - 3ad(a - d) + a^3 + a^3 + d^3 + 3ad(a + d) = 288

a^3 - d^3 - 3a^2d + 3ad^2 + a^3 + d^3 + 3a^2d + 3ad^2 = 288

3a^3 + 6ad^2 = 288

a^3 + 2ad^2 = 96

(4)^3 + 2(4)d^2 = 96

64 + 8d^2 = 96

8d^2 = 96 - 64

8d^2 = 32

d^2 = 32/8

d^2 = 4

d = +2, -2.

When d = +2:

The numbers are =  4 - 2,4,4 + 2

                              = 2,4,6


When d = -2.

The numbers are = 4 + 2,4,4 - 2

                              = 6,4,2.


Hence the required numbers are 2,4,6.


Hope this helps!

Answer 2

Refer the attached picture.

As you asked, I send the Sample Paper of Maths, here. Check it.