Right option is (c) \(\frac {b^2-2ca}{a^2}\)
For explanation: β and γ are the zeros of the polynomial ax^3 + bx^2 + cx + d
So, α + β + γ = \(\frac {-b}{a}\)
αβ + βγ + γα = \(\frac {c}{a}\)
Now, α^2 + β^2 + γ^2 = (α + β + γ)^2 – 2(αβ + βγ + γα)
α^2 + β^2 + γ^2 = \((\frac {-b}{a})\)^2 – 2\((\frac {c}{a}) = \frac {b^2}{a^2}\) – 2 \(\frac {c}{a} = \frac {b^2-2ca}{a^2}\)