Prove that:

Solution:

Let LHS $=\Delta=\mid a^{2}+1 \quad a b \quad a c$

$a b \quad b^{2}+1 \quad b c$

$c a \quad c b \quad c^{2}+1 \mid$

$=(a b c) \mid a+\frac{1}{a} \quad b \quad c$

$a \quad b+\frac{1}{b} \quad c$

$\begin{array}{llll}a & b & c+\frac{1}{c}\end{array} \quad$ [Taking out $a, b$ and $c$ common from $R_{1}, R_{2}$ and $R_{3}$ ]

$=(a b c) \mid a+\frac{1}{a} \quad b \quad c$

$\begin{array}{lll}-\frac{1}{a} & \frac{1}{b} & 0\end{array}$

$\begin{array}{lll}-\frac{1}{a} & 0 & \frac{1}{c}\end{array} \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

$=(a b c)\left(\frac{1}{a b c}\right) \mid a^{2}+1 \quad b^{2} \quad c^{2}$

$\begin{array}{lll}-1 & 1 & 0\end{array}$

$-1 \quad 0 \quad 1 \mid \quad\left[\right.$ Applying $C_{1} \rightarrow a C_{1}, C_{2} \rightarrow b C_{2}$ and $\left.C_{3} \rightarrow c C_{3}\right]$

$=\mid a^{2}+1 \quad b^{2} \quad c^{2}$

$\begin{array}{lll}-1 & 1 & 0 \\ -1 & 0 & 1\end{array}$

$=(-1) \mid b^{2} \quad c^{2}$

1 $0|+(1)| a^{2}+1 \quad b^{2}$    [Expanding along $\mathrm{R}_{3}$ ]

$=(-1)\left(-c^{2}\right)+\left(a^{2}+1+b^{2}\right)$

$=\left(a^{2}+1+b^{2}+c^{2}\right)$

$=\left(a^{2}+b^{2}+c^{2}+1\right)$

$=$ RHS