Given that $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$


$\begin{aligned} & \text { Now, }\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right) \\ & \qquad=\left(1+\alpha^2\right)\left(1+\beta^2+\gamma^2+(\beta \gamma)^2\right) \\ & =1+\beta^2+\gamma^2+(\beta \gamma)^2+\alpha^2+(\alpha \beta)^2+(\alpha \gamma)^2+(\alpha \beta \gamma)^2 \\ & =1+\left(\alpha^2+\beta^2+\gamma^2\right)+\left((\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2\right. \\ & =1+\left[(\alpha \beta \gamma)^2\right. \\ & +\left[(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)\right]+(\alpha \beta \gamma(\alpha+\beta+\gamma)]+(\alpha \beta \gamma)^2\end{aligned}$
From Eq. $(i),(i i)$ and (iii), we get
$$
\begin{aligned}
& =1+\left[p^2-2 q\right]+\left[q^2-2 r p\right]+r^2 \\
& =1+p^2-2 q+q^2-2 r p+r^2 \\
& =(q-1)^2+(r-p)^2
\end{aligned}
$$