If $\alpha ,\beta ,\gamma$ be the roots of x³ + x + 2 = 0, find an equation whose roots are $(\alpha -\beta {)}^2,(\beta -\gamma {)}^2,(\gamma -\alpha {)}^2$

Given equation is x³ + x + 2 = 0 Thus, $\alpha +\beta +\gamma =0,\alpha \beta +\alpha \gamma +\beta \gamma =1,\alpha \beta \gamma =-2$

Let 

$\begin{array}{l}y=(\alpha -\beta {)}^2={\alpha }^2+{\beta }^2-2\alpha \beta \\ =(\alpha +\beta {)}^2-2\alpha \beta -2\alpha \beta \\ ={\gamma }^2-4\alpha \beta \\ ={\gamma }^2-\frac{4\alpha \beta \gamma }{\gamma }\\ ={\gamma }^2+\frac{8}{\gamma }\\ =\frac{{\gamma }^3+8}{\gamma }=\frac{6-\gamma }{\gamma }=\frac{6}{\gamma }-1\end{array}$

$\begin{array}{r}\Rightarrow y=\frac{6}{\gamma }-1\\ \Rightarrow \gamma =\frac{6}{y+1}\end{array}$

Since $\gamma$ is a root of the given equation, so

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

Hence, the required equation is

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