If \(\alpha, \beta\) are roots of equation \(a x^{2}+b x+c=0\) then value of \((\alpha a+b)^{-2}+(\beta a+b)^{-2}\) is (a) \(\left[\left(b^{2}-4 a c\right) /\left(a^{2} c^{2}\right)\right]\) (b) \(\left[\left(b^{2}-a c\right) /\left(a^{2} c^{2}\right)\right]\) (c) \(\left[\left(b^{2}-2 a c\right) /\left(a^{2} c^{2}\right)\right]\) (d) \(\left[\left(b^{2}+2 a c\right) /\left(a^{2} c^{2}\right)\right]\)

For the quadratic equation \(ax^2 + bx + c = 0\), the sum of roots (α + β) and product of roots (αβ) can be expressed as follows: Sum of roots: \(\alpha + \beta = \frac{-b}{a}\) Product of roots: \(\alpha \beta = \frac{c}{a}\)

Substitute \(\alpha = \frac{-b - \beta a}{a}\) in \((\alpha a+b)^{-2}\). \((\alpha a+b)^{-2} = \left(\frac{(-b - \beta a)a + b}{a}\right)^{-2}\) \((\alpha a+b)^{-2} = \left(\frac{(\beta a^2 - b^2)}{a^2}\right)^{-2}\) Similarly, substitute \(\beta = \frac{-b - \alpha a}{a}\) in \((\beta a+b)^{-2}\). \((\beta a+b)^{-2} = \left(\frac{(-b - \alpha a)a + b}{a}\right)^{-2}\) \((\beta a+b)^{-2} = \left(\frac{(\alpha a^2 - b^2)}{a^2}\right)^{-2}\) Now the expression becomes: \((\alpha a+b)^{-2}+(\beta a+b)^{-2} = \left(\frac{(\beta a^2 - b^2)}{a^2}\right)^{-2} + \left(\frac{(\alpha a^2 - b^2)}{a^2}\right)^{-2}\)

Add the fractions and simplify the expression: \(= \left(\frac{1}{(\beta a^2 - b^2)^2}\right) + \left(\frac{1}{(\alpha a^2 - b^2)^2}\right)\) \(= \frac{(\alpha a^2 - b^2)^2+(\beta a^2 - b^2)^2}{(\alpha a^2 - b^2)^2(\beta a^2 - b^2)^2}\)

Now, remember that \(\alpha \beta = \frac{c}{a}\), so \(\alpha a^2\beta - \alpha b^2 = c\). Substitute this back into the expression: \(= \frac{(\alpha a^2\beta - \alpha b^2 + \beta a^2\alpha - \beta b^2)^2}{(\alpha a^2 - b^2)^2(\beta a^2 - b^2)^2}\) After simplification: \(= \frac{(b^2(\alpha + \beta) - 2 \alpha \beta a^2)^2}{(\alpha a^2 - b^2)^2(\beta a^2 - b^2)^2}\) Now, we'll replace \(\alpha + \beta\) with \(\frac{-b}{a}\) and \(\alpha \beta\) with \(\frac{c}{a}\): \(= \frac{(b^2\left(\frac{-b}{a}\right) - 2\left(\frac{c}{a}\right)a^2)^2}{(\alpha a^2 - b^2)^2(\beta a^2 - b^2)^2}\) Simplifying and canceling terms: \(= \frac{\left(b^2 - 2ac\right)^2}{a^2c^2}\) So the correct answer is (c) \(\left[\left(b^{2}-2 a c\right) /\left(a^{2} c^{2}\right)\right]\).