Given that \(\alpha\) and \(\beta\) are zeros of the quadratic polynomial f(x) = ax² + bx + c.

\(\because\) sum of zeros is \(\alpha\) + \(\beta\) \(=\frac{-b}{a}\).

And product of the zeros is \(\alpha \beta\) \(=\frac{c}{a}\).

Now, \(\frac{(\alpha+\beta)^2}{\alpha\beta}\) \(=\frac{\alpha^2}{\alpha\beta}+\frac{\beta^2}{\alpha\beta}+\frac{2\alpha\beta}{\alpha\beta}\) 

\(=\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2\)

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

\(=\frac{b^2}{ac}-2\) \(=\frac{b^2-2ac}{ac}\)

\(\Rightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) \(=\frac{b^2-2ac}{ac}\) ...........(1)

And \(\frac{(\alpha+\beta)^3}{\alpha\beta}\) \(=\frac{\alpha^3+\beta^3+3\alpha^2\beta+3\alpha\beta^2}{\alpha\beta}\)  

\(\Big(\because (a+b)^3\) \(=a^3+b^3+3a^2b + 3ab^2\Big)\)

\(=\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}+3\alpha+3\beta\)

\(=\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}+3(\alpha+\beta)\)

\(\Rightarrow \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\) \(=\frac{(\alpha+\beta)^3}{\alpha\beta}\) \(-3(\alpha+\beta)\)

\(=(\alpha+\beta)\left(\frac{(\alpha+\beta)^2}{\alpha\beta}-3\right)\)

\(\Rightarrow \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\) \(=\frac{-b}{a}\left(\frac{\left(\frac{-b}{a}\right)^2}{\frac{c}{a}}-3\right)\)  

(By putting values of \(\alpha \) + \(\beta\) and \(\alpha \)\(\beta\))

\(=\frac{-b}{a}\left(\frac{b^2}{ac}-3\right)\) ..........(2)

Now, \(a\times \left(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right)\) \(+b\times \left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)\)  

\(=a\left(\frac{-b}{a}\times \frac{b^2-3ac}{ac}\right)\)\(+b\times \left(\frac{b^2-2ac}{ac}\right)\)  (From equation (1) and (2))

\(=\frac{-b^3+3abc}{ac}+\frac{b^3-2abc}{ac}\)

\(=\frac{-b^3+3abc+b^3-2abc}{ac}\)

\(=\frac{abc}{ac}\)

\(=b\).