If a,b,c are in arithmetic progression, then the value of (a+2b-c)(2b+c-a)(a+2b+c)
Finding the value of :
Given a,b,c are in A.P
So 2b=a+c
and (a+2b-c)(2b+c-a)(a+2b+c)....(i)
we get (a+a+c-c)(a+c+c-a)(a+a+c+c)
=(2a)(2c)(2a+2c)=8ac(a+c)=8ac(2b)=16abc
Therefore (a+2b-c)(a+2b+c)(2b+c-a)=16abc
Hence, the value of (a+2b-c)(2b+c-a)(a+2b+c) is 16abc