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Question
\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]
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Solution
\[∆ = \begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix}\]
\[ = \begin{vmatrix}0 & - 2c & - 2b \\ b & c + a & b \\ c & c & a + b\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 - ( R_2 + R_3 ) \right]\]
\[ = \begin{vmatrix}0 & - 2c & - 2b \\ b & c + a - b & 0 \\ c & 0 & a + b - c\end{vmatrix} \left[\text{ Applying }C_2 \text{ to }C_2 - C_1\text{ and }C_3 \text{ to }C_3 - C_1 \right]\]
\[ = 0\begin{vmatrix}c + a - b & 0 \\ 0 & a + b - c\end{vmatrix} - ( - 2c)\begin{vmatrix}b & 0 \\ c & a + b - c\end{vmatrix} - 2b\begin{vmatrix}b & c + a - b \\ c & 0\end{vmatrix} \left[\text{ Expanding along }R_1 \right]\]
\[ = 2c[b(a + b - c) - 0] - 2b[0 - c(c + a - b)]\]
\[ = 2bc[a + b - c] - 2bc[b - c - a]\]
\[ = 2bc[(a + b - c) - (b - c - a)]\]
\[ = 4abc\]
Hence proved.
