If a,b,c are in A.P, then prove that -

(a+2b-c)(2b+c-a)(c+a-b)=4abc

If a, b, a are in A.P.    ....we have that......

a = a

b = a + d

c = a + 2d

(a+2b-c)(2b+c-a)(c+a-b)=4abc

[ a + 2(a + d) - (a + 2d] [ 2(a + d) + (a + 2d) - a ] [ (a + 2d) + a - (a + d) ]  = 4[a][a + d] [a + 2d]

[ 2a] [ 2a + 4d] [ a + d ]  =  4a [a^2 + 3ad + 2d^2]

[ 4a^2 + 8ad] [ a + d ]  =   4a^3  + 12a^2d + 8ad^2

4a^3 + 8a^2d + 4a^2d + 8ad^2  = 4a^3  + 12a^2d + 8ad^2

4a^3 + 12a^2d + 8ad^2  = 4a^3 + 12a^2d + 8ad^2