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Q. Let $\cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\alpha-\gamma)=3$, where $\alpha, \beta, \gamma \in[0, \pi]$. If $\alpha, \beta, \gamma$ are roots of $x ^3+ px ^2+ qx + r =0$ then which of the following must be true?

Complex Numbers and Quadratic Equations

Solution:

$ \cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\alpha-\gamma)=3$
$\Rightarrow \alpha-\beta=0=\beta-\gamma=\alpha-\gamma$
$\Rightarrow \alpha=\beta=\gamma$
For cubic equation, $ x ^3+ px ^2+ qx + r$
$\Rightarrow 3 \alpha=- p $ ....(1)
$\Rightarrow 3 \alpha^2= q $....(2)
$\Rightarrow \alpha^3=- r $....(3)
$\text { From (3), }$
$\alpha \cdot \alpha^2=- r \Rightarrow \frac{- p }{3} \cdot \frac{ q }{3}=- r $
$\Rightarrow pq =9 r$