If a,b,c are in ap then the value of (a+2b-c)(2b+c-a)(a+2b+c)

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Solution

If a,b,c are in AP then

2b=c+a..........(i)

now,

( a+2b-c)(2b+c-a)(c+a+2b)

= ( a+c+a-c)(c+a+c-a)(2b+2b) [from(i)]

= (2a) (2c) (4b)

= 16abc

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