If $α, β, γ$ be the roots of $x^{3} + x + 2 = 0$ , find an equation whose root are $(α - β)^{2}, (β - γ)^{2}, (γ - α)^{2}$

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$x^{4} + 6 x^{6} + 9 x + 112 = 0$

$x^{3} + 6 x^{2} + 15 x + 60 = 0$

$x^{3} + 6 x^{2} + 9 x + 112 = 0$

answer is D.

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Detailed Solution

Given equation is $x^{3} + x + 2 = 0$ Thus, $α + β + γ = 0, α β + α γ + β γ = 1, α β γ = - 2$

Let

$\begin{array}{l} y = (α - β)^{2} = α^{2} + β^{2} - 2 α β \\ = (α + β)^{2} - 2 α β - 2 α β \\ = γ^{2} - 4 α β \\ = γ^{2} - \frac{4 α β γ}{γ} \\ = γ^{2} + \frac{8}{γ} \\ = \frac{γ^{3} + 8}{γ} = \frac{6 - γ}{γ} = \frac{6}{γ} - 1 \end{array}$

$\begin{aligned} \Rightarrow y = \frac{6}{γ} - 1 \\ \Rightarrow γ = \frac{6}{y + 1} \end{aligned}$

Since $γ$ is a root of the given equation, so

$\begin{array}{l} γ^{3} + γ + 2 = 0 \\ \Rightarrow {(\frac{6}{y + 1})}^{3} + (\frac{6}{y + 1}) + 2 = 0 \end{array}$

Hence, the required equation is

$\begin{array}{l} \Rightarrow {(\frac{6}{y + 1})}^{3} + (\frac{6}{y + 1}) + 2 = 0 \\ \Rightarrow 2 (x + 1)^{3} + 6 (x + 1)^{2} + 216 = 0 \\ \Rightarrow x^{3} + 6 x^{2} + 9 x + 112 = 0 \end{array}$