If $a^2+b^2+c^2=-2$  and $f\left(x\right)=$  $\left|\begin{array}{ccc}1+a^{2}x& (1+b^2)x& (1+c^2)x\\ (1+a^2)x& 1+b^{2}x& (1+c^2)x\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$, then f(x) is a polynomial of degree

$\begin{array}{l}f\left(x\right)=\left|\begin{array}{ccc}1+(a^2+b^2+c^2+2)x& (1+b^2)x& (1+c^2)x\\ 1+(a^2+b^2+c^2+2)x& 1+b^{2}x& (1+c^2)x\\ 1+(a^2+b^2+c^2+2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|\\ \end{array}$

Applying  $C_1\to C_1+C_2+C_3$

$\left[\because a^2+b^2+c^2=-2,\text{ given }\right]$

$\left|\begin{array}{ccc}1& \left(1+b^2\right)x& \left(1+c^2\right)x\\ 1& 1+b^{2}x& \left(1+c^2\right)x\\ 1& \left(1+b^2\right)x& 1+c^{2}x\end{array}\right|$

$\text{ Applying }{R}_1\to R_1-R_3,R_2\to R_2-R_3\text{, we get }$

$\left|\begin{array}{ccc}0& 0& x-1\\ 0& 1-x& x-1\\ 1& \left(1+b^2\right)\times & 1+c^{2}x\end{array}\right|$

$=\left|\begin{array}{cc}0& x-1\\ 1-x& x-1\end{array}\right|=(x-1{)}^2$