Sol:
Given:
\(P(x) = ax^{3} + bx^{2} + cx + d\) …………………..[1]Where \(a\neq 0\) is a cubic polynomial
If α,β,γ are the zeroes of the polynomial
then
\(p(x) = (x – \alpha)(x – \beta) (x – \gamma)\)⇒ \(p(x) = x^{3} – (\alpha +\beta +\gamma) x^{2} + (\alpha\beta + \beta\gamma + \gamma\alpha) x – (\alpha +\beta +\gamma)\) …………………..[2]
Now both the equations (1) and (2) are same. By comparing the coefficients, we get,
⇒ \(\frac{a}{1} = \frac{b}{-\alpha -\beta -\gamma} = \frac{c}{\alpha \beta + \beta \gamma + \alpha}=\frac{d}{-\alpha\beta\gamma}\)
On solving we get,
⇒ \(\alpha +\beta +\gamma = \frac{-b}{a}\) = Sum of the roots
⇒ \(\alpha\beta +\beta\gamma +\gamma\alpha = \frac{c}{a}\)
⇒ \(\alpha\beta\gamma = \frac{-d}{a}\) = Product of the roots
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