If  $\alpha ,\beta ,\gamma$  are the roots of  $x^3+p{x}^2+qx+r=0$  then the value of  $(1+{\alpha }^2)(1+{\beta }^2)(1+{\gamma }^2)=$  

Take  $\alpha =1\beta =2\gamma =3$
 $$\begin{gathered} (x-1)(x-2)(x-3)=0 \\ {x}^3-6x^2+11x-6=0 \\ \therefore p=-6q=11r=6 \\ (1+{\alpha }^2)(1+{\beta }^2)(1+{\gamma }^2)=100 \\ {(r-p)}^2+{(q-1)}^2=0+100=100 \end{gathered}$$