If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = a x^2 + b x + c$, evaluate:
(i) $\alpha^3 + \beta^3$ [2.5 Marks]
(ii) $\dfrac{1}{\alpha^3} + \dfrac{1}{\beta^3}$ [2.5 Marks]
Step-by-Step Solution
Key Concept: $\alpha + \beta = -b/a, \alpha \beta = c/a$.<br>(i) $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta) = \dfrac{3abc - b^3}{a^3}$.<br>(ii) $\dfrac{1}{\alpha^3} + \dfrac{1}{\beta^3} = \dfrac{\alpha^3 + \beta^3}{\alpha^3 \beta^3} = \dfrac{3abc - b^3}{c^3}$.
(i) $\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha \beta] = \left(-\dfrac{b}{a}\right)\left[\dfrac{b^2}{a^2} - \dfrac{3c}{a}\right] = \dfrac{-b^3 + 3abc}{a^3} = \dfrac{3abc - b^3}{a^3}$. [2.5 Marks]
(ii) $\dfrac{1}{\alpha^3} + \dfrac{1}{\beta^3} = \dfrac{\alpha^3 + \beta^3}{\alpha^3 \beta^3} = \dfrac{\dfrac{3abc - b^3}{a^3}}{\dfrac{c^3}{a^3}} = \dfrac{3abc - b^3}{c^3}$. [2.5 Marks]
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🎯 Official CBSE Marking Scheme:
Part (i) Deriving $\alpha^3 + \beta^3 = (3abc - b^3)/a^3$: 2.5 Marks
Part (ii) Deriving $\dfrac{1}{\alpha^3} + \dfrac{1}{\beta^3} = (3abc - b^3)/c^3$: 2.5 Marks
Correct Answer: