If $\alpha, \beta, y$ are the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$, then $\alpha^{2}+\beta^{2}+y^{2}=$
(a) $\frac{b^{2}-a c}{a^{2}}$
(b) $\frac{b^{2}-2 a c}{a}$
(c) $\frac{b^{2}+2 a c}{b^{2}}$
(d) $\frac{b^{2}-2 a c}{a^{2}}$
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)