Prove that:

Solution:

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

$\begin{array}{lcc}a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}$

$\Delta=a b c \mid a \quad c \quad a+c$

$\begin{array}{lcc}a+b & b & a \\ b & b+c & c \mid\end{array}$ [Taking out $a, b$ and $c$ common from $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ ]

$\begin{array}{llll}=a b c & \mid a & c & 0 \\ a+b & b & -2 b & \end{array}$

$b \quad b+c \quad-2 b \mid \quad\left[\right.$ Applying $\left.\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{2}-\mathrm{C}_{1}\right]$

$=(a b c)(-2 b) \mid a \quad c \quad 0$

$\begin{array}{lll}\mathrm{a} & -\mathrm{c} & 0\end{array}$

$\mathrm{b} \quad \mathrm{b}+\mathrm{c} \quad 1 \mid \quad$ [Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$ ]

$=(\mathrm{abc})(-2 \mathrm{~b}) \times 1 \mid \mathrm{a} \quad \mathrm{c}$

a $\quad-c \mid \quad\left[\right.$ Exp anding along $\left.\mathrm{C}_{3}\right]$

$\begin{array}{lll}\mathrm{a} & -\mathrm{c} \mid & {\left[\text { Exp anding along } \mathrm{C}_{3}\right]}\end{array}$

$=(\mathrm{abc})(-2 \mathrm{~b})(-2 \mathrm{ac})$

$=4 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}$

$=\mathrm{RHS}$