if a^2(b+c), b^2(c+a), c^2(a+b) are in A.P, prove that either a,b,c are in A.P or ab+bc+ca=0

Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.
So, b²(c+a) - a²(b+c) = c²(a+b) - b²(c+a)
==> b²c + b²a - a²b - a²c = c²a + c²b - b²c - b²a
==>(b²c - a²c) + (b²a - a²b) = (c²a - b²a) + (c²b - b²c)
==>c(b² - a²) + ab(b-a) = a(c² - b²) + bc(c - b)
==>(b-a) [ c(b+a) + ab ] = (c-b) [ a(c+b) + bc ]
==> (b-a)( ab + bc ca ) = (c-b)( ab + bc + ca )

∴ Either : ab + bc + ca = 0
Or : b - a = c - b, i.e., a, b, c are in A.P.

Thanks, Vj