Explanation

(a)
Applying $C_{1} \rightarrow C_{1} + C_{2} + C_{3},\$we get
$f(x) = \left| \begin{matrix}
1 + a^{2}x + x + xb^{2} + x + c^{2}x & \left( 1 + b^{2} \right)x & \left( 1 + c^{2} \right)x \\
x + a^{2}x + 1 + b^{2}x + x + c^{2}x & \left( 1 + b^{2}x \right) & \left( 1 + c^{2} \right)x \\
x + a^{2}x + x + b^{2}x + 1 + c^{2}x & \left( 1 + b^{2} \right)x & \left( 1 + c^{2}x \right)
\end{matrix} \right|$
$= \left| \begin{matrix}
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{matrix} \right|\ \$
$\lbrack\because a^{2} + b^{2} + c^{2} + 2 = 0\rbrack$
Applying $R_{1} \rightarrow R_{1} - R_{3},\ \ R_{2} \rightarrow R_{2} - R_{3}$
$= \left| \begin{matrix}
0 & 0 & x - 1 \\
0 & 1 - x & x - 1 \\
1 & \left( 1 + b^{2} \right)x & 1 + c^{2}x
\end{matrix} \right| = 1\lbrack 0 - (x - 1)(1 - x)\rbrack$
$\mathbf{=}(x - 1)^{2}$
$\Rightarrow f(x)$ is a polynomial of degree 2