Question

If $a^2(b+c)$, $b^2(c+a)$, $c^2(a+b)$ are in A.P., then value of $ab+bc+ca$ is

Answer 1

$a^2(b+c)+b^2(c+a)+c^2(a+b)$ are in AP

$\therefore b^2(c+a)-a^2(b+c)=c^2(a+b)-b^2(c+a)$

$b^{2}c+b^{2}a-a^{2}b-a^{2}c=c^{2}a+c^{2}b-b^{2}c-b^{2}a$

$(b^{2}c-a^{2}c)+(b^{2}a-a^{2}b)=(c^{2}a-b^{2}a)+(c^{2}b-b^{2}c)$

$c(b^2-a^2)+ab(b-a)=a(c^2-b^2)+bc(c-b)$

$(b-a)\left[c\right(b+a)+ab]=(c-b)\left[a\right(c+b)+bc]$

$(b-a)(ab+bcca)=(c-b)(ab+bc+ca)$

$ab+bc+ca=0$

Or : $b-a=c-b$, i.e., a, b, c are in A.P.