SOLUTION:

Let the roots of the equation be α and β. Then

α + β = _{1/}α² + 1/β²

or, α + β = (α²+β²) / α²β²

or, α + β = ( α + β)² - 2αβ / α²β²

Putting the values of α + β and αβ we get

-b/a =( (b²/a²) - 2c/a) / (c²/a²)

or, -b/a = (b²- 2ac)/ c²

or, -bc² = ab² - 2ca²

or bc² + ab² = 2ca²

Hence bc² , ca², ab² are in A.P.