If $α, β, γ$ are the roots of $x^{3} - 6 x^{2} + 11 x - 6 = 0$ then the equation having the roots $α^{2} + β^{2}, β^{2} + γ^{2}$ and $γ^{2} + α^{2}$ is

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$x^{3} - 28 x^{2} + 245 x - 650 = 0$

$x^{3} - 28 x^{2} + 245 x + 650 = 0$

$x^{3} + 28 x^{2} - 245 x - 650 = 0$

$x^{3} + 28 x^{2} + 245 x - 650 = 0$

answer is A.

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Detailed Solution

$\begin{array}{l} I f α, β, γ a r e r o o t s o f x^{3} - 6 x^{2} + 11 x - 6 = 0 \\ B y t r a i l a n d e r r o r x = 1 i s r o o t \end{array}$

$\begin{array}{l} ∴ (x - 1) (x^{2} - 5 x + 6) = 0 \\ \Rightarrow (x - 1) (x - 2) (x - 3) = 0 \\ ∴ x = 1, 2, 3 \\ ∴ α = 1, β = 2, γ = 3 \\ a = α^{2} + β^{2} = 1 + 4 = 5 \\ b = β^{2} + γ^{2} = 4 + 9 = 13 \\ c = γ^{2} + α^{2} = 9 + 1 = 10 \\ Re q u i r e d E q u a t i o n i s x^{3} - (a + b + c) x^{2} + (a b + b c + c a) x + a b c = 0 \\ \Rightarrow x^{3} - 28 x^{2} + 245 x - 650 = 0 \end{array}$