If $α, β, γ$ are the roots of $x^{3} + {px}^{2} + qx + r = 0$ , then the value of $(1 + α^{2}) (1 + β^{2}) (1 + γ^{2})$ is

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${(r + p)}^{2} + {(q + 1)}^{2}$

${(r - p)}^{2} + {(q - 1)}^{2}$

${(1 + p)}^{2} + {(1 + q)}^{2}$

${(r - p)}^{2} + {(r - q)}^{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} I f α, β, γ a r e r o o t s o f x^{3} + p x^{2} + q x + r = 0 \\ T h e n α + β + γ = - p, α β + β γ + γ α = q, α β γ = - r \\ ∴ (1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) = 1 + (α^{2} + β^{2} + γ^{2}) + (α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2}) + α^{2} β^{2} γ^{2} \\ = 1 + {(α + β + γ)}^{2} - 2 (α β + β γ + γ α) + {(α β + β γ + γ α)}^{2} - 2 α β γ (α + β + γ) + {(α β γ)}^{2} \\ = 1 + p^{2} - 2 q + q^{2} - 2 p r + r^{2} \\ = (p^{2} + r^{2} - 2 p r) + (q^{2} - 2 q + 1) \\ = {(p - r)}^{2} + {(q - 1)}^{2} \end{array}$