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Question
Prove that :
Solution
\[\text{ Let LHS }= \Delta = \begin{vmatrix} a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac\\ab & b^2 + bc & c^2 \end{vmatrix}\]
\[\Delta = abc \begin{vmatrix} a & c & a + c\\a + b & b & a\\b & b + c & c \end{vmatrix} \left[\text{ Taking out a, b and c common from }C_1 , C_2\text{ and }C_3 \right]\]
\[ = abc \begin{vmatrix} a & c & 0\\a + b & b & - 2b\\b & b + c & - 2b \end{vmatrix} \left[\text{ Applying }C_3 \to C_3 - C_2 - C_1 \right]\]
\[ = \left( abc \right)\left( - 2b \right) \begin{vmatrix} a & c & 0\\a + b & b & 1\\b & b + c & 1 \end{vmatrix}\left[\text{ Taking ( - 2b) common from }C_3 \right]\]
\[ = \left( abc \right)\left( - 2b \right) \begin{vmatrix} a & c & 0\\a & - c & 0\\b & b + c & 1 \end{vmatrix} \left[\text{ Applying }R_2 \to \hspace{0.167em} R_2 - R_3 \right]\]
\[ = \left( abc \right)\left( - 2b \right) \times 1\begin{vmatrix} a & c \\a & - c \end{vmatrix} \left[\text{ Expanding along }C_3 \right]\]
\[ = \left( abc \right)\left( - 2b \right)\left( - 2ac \right)\]
\[ = 4 a^2 b^2 c^2 \]
\[ = RHS\]
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