If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }, \sigma$ are the roots of the equation ${\mathrm{x}}^4+4{\mathrm{x}}^3-6{\mathrm{x}}^2+7\mathrm{x}-9=0$ then the value of $\left(1+{\mathrm{\alpha }}^2\right)\left(1+{\mathrm{\beta }}^2\right)\left(1+{\mathrm{\gamma }}^2\right)\left(1+{\mathrm{\sigma }}^2\right)$ is

Since, $\mathrm{\alpha },\mathrm{\beta }, \mathrm{\gamma }, \mathrm{\sigma }$ are the roots of the given equation, we have

${\mathrm{x}}^4+4{\mathrm{x}}^3-6{\mathrm{x}}^2+7\mathrm{x}-9=(\mathrm{x}-\mathrm{\alpha })(\mathrm{x}-\mathrm{\beta })(\mathrm{x}-\mathrm{\gamma })(\mathrm{x}-\mathrm{\sigma })$

Putting x = i and then x = -i , we get

$\begin{array}{l} 1-4\mathrm{i}+6+7\mathrm{i}-9=(\mathrm{i}-\mathrm{\alpha })(\mathrm{i}-\mathrm{\beta })(\mathrm{i}-\mathrm{\gamma })(\mathrm{i}-\mathrm{\sigma })\\ \mathrm{and} 1+4\mathrm{i}+6-7\mathrm{i}-9=(-\mathrm{i}-\mathrm{\alpha })(-\mathrm{i}-\mathrm{\beta })(-\mathrm{i}-\mathrm{\gamma })(-\mathrm{i}-\mathrm{\sigma })\end{array}$

Multiplying these two equations, we get

$\begin{array}{l} (-2+3\mathrm{i})(-2-3\mathrm{i})=\left(1+{\mathrm{\alpha }}^2\right)\left(1+{\mathrm{\beta }}^2\right)\left(1+{\mathrm{\gamma }}^2\right)\left(1+{\mathrm{\sigma }}^2\right)\\ \mathrm{or} 13=\left(1+{\mathrm{\alpha }}^2\right)\left(1+{\mathrm{\beta }}^2\right)\left(1+{\mathrm{\gamma }}^2\right)\left(1+{\mathrm{\sigma }}^2\right)\end{array}$