If α, β, γ are the zeros of the polynomial f(x) = ax3 + bx2 + cx + d,

Solution:

We have to find the value of $\alpha^{2}+\beta^{2}+\gamma^{2}$

Given $\alpha, \beta, \gamma$ be the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$

$\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$

$=\frac{-b}{a}$

$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{3}}$

$=\frac{c}{a}$

Now

$\alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^{2}-2(\alpha \beta+\beta \gamma+\gamma \alpha)$

$\alpha^{2}+\beta^{2}+\gamma^{2}=\left(\frac{-b}{a}\right)^{2}-2\left(\frac{c}{a}\right)$

$\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a}$

$\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{b^{2}}{a^{2}}-\frac{2 c \times a}{a \times a}$

$\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{b^{2}}{a^{2}}-\frac{2 c a}{a^{2}}$

$\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{b^{2}-2 a c}{a^{2}}$

The value of $\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{b^{2}-2 a c}{a^{2}}$

Hence, the correct choice is (d)