If $\alpha ,\beta ,\gamma$  are the roots of $x^3+x+1=0$  then the equation whose roots are ${\left(\alpha -\beta \right)}^2,{\left(\beta -\gamma \right)}^2,{\left(\gamma -\alpha \right)}^2$  is

$={\left(-\gamma \right)}^2+\frac{4}{\gamma }$

$=\frac{{\gamma }^3+4}{\gamma }$

We know that $\gamma$  is a root

${\gamma }^3+\gamma +1=0$

${\gamma }^3=-\gamma -1$

$=\frac{-\gamma -1+4}{\gamma }$

$=\frac{-\gamma +3}{\gamma }$

$=\frac{3}{\gamma }-1$

$\frac{3}{\gamma }-1=x$

$\frac{3}{\gamma }=x+1$

$\gamma =\frac{3}{x+1}$

$f\left(\frac{3}{x+1}\right)={\left(\frac{3}{x+1}\right)}^3+\frac{3}{x+1}+1$

$=\frac{27}{{\left(x+1\right)}^3}+\frac{3}{\left(x+1\right)}+1$

$={\left(x+1\right)}^3+3{\left(x+1\right)}^2+27$