If $α, β, γ$ are the roots of the equation $x^{3} + {px}^{2} + qx + r = 0$ then $\sum α^{2} (β + γ) =$

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$p^{2} - 2 q$

$- p^{3} + 3 pq - 3 r$

$p^{4} - 4 p^{2} q + 4 pr + 2 q^{2}$

3r – pq

answer is D.

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Detailed Solution

$\begin{array}{l} I f α, β, γ a r e r o o t s x^{3} + p x^{2} + q x + r = 0 \\ T h e n α + β + γ = - p, α β + β γ + γ α = q, α β γ = - r \\ \sum α^{2} (β + γ) = α^{2} (β + γ) + β^{2} (α + γ) + γ^{2} (α + β) \\ = α^{2} (- p - α) + β^{2} (- p - β) + γ^{2} (- p - γ) \\ = - α^{2} p - α^{3} - p β^{2} - β^{3} - p γ^{2} - γ^{3} \\ = - p (α^{2} + β^{2} + γ^{2}) - (α^{3} + β^{3} + γ^{3}) \\ = - p ({(α + β + γ)}^{2} - 2 (α β + β γ + γ α)) - {(α + β + γ)}^{3} + 3 α β γ (α + β + γ) \\ = - p (p^{2} - 2 q) - {(- p)}^{3} + 3 (- r) (- p) \\ = - p^{3} + 2 p q + p^{3} + 3 p r ∵ 2 p q + p^{3} \end{array}$