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

Let's simplify the determinant using column operations: $C_1\to C_1+C_2+C_3$.The first column becomes:$1+a^{2}x+(1+b^2)x+(1+c^2)x=1+(a^2+b^2+c^2+2)x$ Given $a^2+b^2+c^2=-2$, the first column entry becomes:$1+(-2+2)x=1$.The same happens for all entries in $C_1$:
$$f(x)=\left|\begin{array}{ccc}1& (1+b^2)x& (1+c^2)x\\ 1& 1+b^{2}x& (1+c^2)x\\ 1& (1+b^2)x& 1+c^{2}x\end{array}\right|$$
Now apply $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$:
$$f(x)=\left|\begin{array}{ccc}1& (1+b^2)x& (1+c^2)x\\ 0& 1+b^{2}x-(1+b^2)x& 0\\ 0& 0& 1+c^{2}x-(1+c^2)x\end{array}\right|$$
$f(x)=\left|\begin{array}{ccc}1& (1+b^2)x& (1+c^2)x\\ 0& 1-x& 0\\ 0& 0& 1-x\end{array}\right|$ Expanding along the first column:$f(x)=(1-x)(1-x)=(1-x{)}^2=x^2-2x+1$.The degree of the polynomial is 2.