If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of ${\mathrm{x}}^3–{\mathrm{x}}^2+8\mathrm{x}–6=0$ then $(1+{\mathrm{\alpha }}^2)(1+{\mathrm{\beta }}^2)(1+{\mathrm{\gamma }}^2)=$

$\begin{array}{l}If\alpha ,\beta ,\gamma arerootsof{x}^3-x^2+8x-6=0\\ \therefore \alpha +\beta +\gamma =1,\alpha \beta +\beta \gamma +\gamma \alpha =8,\alpha \beta \gamma =6\\ Then(1+{\alpha }^2)(1+{\beta }^2)(1+{\gamma }^2)\\ =1+({\alpha }^2+{\beta }^2+{\gamma }^2)+({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2)+{\alpha }^2{\beta }^2{\gamma }^2\\ =1+{(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )+{(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )+{\left(\alpha \beta \gamma \right)}^2\\ =1+1-2\left(8\right)+8^2-2.6\left(1\right)+6^2\\ =2-16+64-12+36\\ =102-28\\ =74\end{array}$