If `a,b,c` are in AP, than show that `a^(2)(b+c)+b^(2)(c+a)+c^(2)(a+b)=(2)/(9)(a+b+c)^(3)`.

Explanation

`therefore a,b,c` are in AP.
`therefore b=(a+c)/(2)i.e., 2b=a+c " " "…….(i)"`
LHS`=a^(2)(b+c)+b^(2)(c+a)+c^(2)(a+b)`
`=(a^(2)b+a^(2)c)+b^(2)(2b)+(c^(2)a+c^(2)b)`
`=b(a^(2)+c^(2))+ac(a+c)+2b^(3)`
`=b[(a+c)^(2)-2ac]+ac(2b)+2b^(3)`
`=b(a+c)^(2)+2b^(3)=b(2b)^(2)+2b^(3)=6b^(3)`
RHS `=(2)/(9)(a+b+c)^(3)=(2)/(9)(2b+b)^(3)`
`=(2)/(9)xx27b^(3)=6b^(3)`
Hence, LHS=RHS.