Determinant Setup

We need to evaluate the determinant:

$ D = \begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} $

Determinant Row Operations

Apply the row operation $R_1 \rightarrow R_1 + R_2 + R_3$ to simplify the first row:

$ D = \begin{vmatrix} (b+c)+(c+a)+(a+b) & (a-b)+(b-c)+(c-a) & a+b+c \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} = \begin{vmatrix} 2a+2b+2c & 0 & a+b+c \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} $

Factor out $(a+b+c)$ from the first row:

$ D = (a+b+c) \begin{vmatrix} 2 & 0 & 1 \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} $

Determinant Column Operations

Apply the column operation $C_1 \rightarrow C_1 - 2C_3$ to create zeros in the first row:

$ D = (a+b+c) \begin{vmatrix} 2 - 2(1) & 0 & 1 \\ (c+a) - 2b & b-c & b \\ (a+b) - 2c & c-a & c \end{vmatrix} = (a+b+c) \begin{vmatrix} 0 & 0 & 1 \\ c+a-2b & b-c & b \\ a+b-2c & c-a & c \end{vmatrix} $

Expand the determinant along the first row ($R_1$):

$ D = (a+b+c) \left( 0 \cdot C_{11} + 0 \cdot C_{12} + 1 \cdot C_{13} \right) $

Where $C_{13}$ is the determinant of the remaining 2x2 matrix:

$ C_{13} = \begin{vmatrix} c+a-2b & b-c \\ a+b-2c & c-a \end{vmatrix} $

Calculate $C_{13}$:

$ C_{13} = (c+a-2b)(c-a) - (b-c)(a+b-2c) $

Expand and simplify:

$ = (c^2 - ac + ac - a^2 - 2bc + 2ab) - (ab + b^2 - 2bc - ac - bc + 2c^2) $

$ = (c^2 - a^2 - 2bc + 2ab) - (ab + b^2 - 3bc - ac + 2c^2) $

$ = c^2 - a^2 - 2bc + 2ab - ab - b^2 + 3bc + ac - 2c^2 $

$ = -a^2 - b^2 - c^2 + ab + bc + ac $

Determinant Final Simplification

Substitute $C_{13}$ back into the determinant expression:

$ D = (a+b+c)(-a^2 - b^2 - c^2 + ab + bc + ac) $

Rearrange the terms:

$ D = -(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac) $

Using the algebraic identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac)$, we get:

$ D = -(a^3 + b^3 + c^3 - 3abc) $

$ D = 3abc - a^3 - b^3 - c^3 $