By using properties of determinants,show that:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

(N/A) Let $\Delta = \left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|$.
Taking out common factors $a, b,$ and $c$ from $R_{1}, R_{2},$ and $R_{3}$ respectively,we get:
$\Delta = a b c \left|\begin{array}{ccc}a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c}\end{array}\right|$.
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we get:
$\Delta = a b c \left|\begin{array}{ccc}a+\frac{1}{a} & b & c \\ -\frac{1}{a} & \frac{1}{b} & 0 \\ -\frac{1}{a} & 0 & \frac{1}{c}\end{array}\right|$.
Applying $C_{1} \rightarrow a C_{1}, C_{2} \rightarrow b C_{2},$ and $C_{3} \rightarrow c C_{3},$ we get:
$\Delta = a b c \times \frac{1}{a b c} \left|\begin{array}{ccc}a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right| = \left|\begin{array}{ccc}a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$.
Expanding along $R_{3},$ we get:
$\Delta = -1 \left|\begin{array}{cc}b^{2} & c^{2} \\ 1 & 0\end{array}\right| + 1 \left|\begin{array}{cc}a^{2}+1 & b^{2} \\ -1 & 1\end{array}\right|$.
$\Delta = -1(0 - c^{2}) + 1(a^{2} + 1 + b^{2}) = c^{2} + a^{2} + 1 + b^{2} = 1 + a^{2} + b^{2} + c^{2}$.
Hence,the result is proved.