Question

If $a^2(b+c),b^2(c+a),c^2(a+b)$ are in AP, show that either a,b,c are in AP or ab+bc+ca=0

Answer 1

If $x,y,z$ are in A.P. , $xy=z-y$

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

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

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

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

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

$(b-a)(bc+ac+ab)=(c-b)(ac+bc+ab)$

Either $ab+bc+ac=0,$

$b-a=c-b$

$\therefore a,b,c$ are in A.P.